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## Physics Work Problems for High Schools

In this tutorial, we want to practice some problems on work in physics. All these questions are easy and helpful for your high school homework.

## Work Problems: Constant Force

Problem (1): A constant force of 1200 N is required to push a car along a straight line. A person displaces the car by 45 m. How much work is done by the person?

Solution : If a constant force $F$ acts on an object over a distance of $d$, and $F$ is parallel to $d$, then the work done by force $F$ is the product of the force times distance.

In this case, a force of $1200\,{\rm N}$ displaces the car $45\,{\rm m}$. The pushing force is parallel to the displacement. So, the work done by the person is equal to $W=Fd=1200\times 45=54000\,{\rm J}$ The SI unit of work is the joule, ${\rm J}$.

Problem (2): You lift a book of mass $2\,{\rm kg}$ at constant speed straight upward a distance of $2\,{\rm m}$. How much work is done during this lifting by you?

Solution : The force you apply to lift the book must be balanced with the book's weight. So, the exerted force on the book is $F=mg=2\times 10=20\quad{\rm N}$ The book is lifted 2 meters vertically. The force and displacement are both parallel to each other, so the work done by the person is the product of them. $W=Fd=20\times 2=40\quad {\rm J}$

Problem (3): A force of $F=20\,{\rm N}$ at an angle of $37^\circ$ is applied to a 3-kg object initially at rest. The object has displaced a distance of $25\,{\rm m}$ over a frictionless horizontal table. Determine the work done by  (a) The applied force (b) The normal force exerted by the table (c) The force of gravity

Solution : In this problem, the force makes an angle with the displacement. In such cases, we should use the work formula $W=Fd\cos\theta$ where $\theta$ is the angle between force $F$ and displacement. To this object, an external force $F$, normal force $F_N$, and a gravity force $w=mg$ are applied.

(a) Using vector decompositions, the component of the force parallel to the displacement is found to be $F_{\parallel}=F\cos \theta$. Thus, the product of this component parallel to the displacement times the magnitude of displacement gives us the work done by external force $F$ as below \begin{align*} W_F&=\underbrace{F\cos\theta}_{F_{\parallel}}d\\\\ &=(20\times \cos 37^\circ)(25)\\\\&=400\quad {\rm J}\end{align*}  (b) Now, we want to find the work done by the normal force. But let's define what the normal force is.

In physics, ''normal'' means perpendicular. When an object is in contact with a surface, a contact force is exerted on the object. The component of the contact force perpendicular to the surface is called the normal force.

Thus, by definition, the normal force is always perpendicular to the displacement. So, the angle between $F_N$ and displacement $d$ is $90^\circ$. Hence, the work done by the normal force is determined to be $W_N=F_N d\cos\theta=(30)(25)\cos 90^\circ=0$ (c) The weight of the object is the same as the force of gravity. This force applies to the object vertically downward, and the displacement of the object is horizontal. So, again, the angle between these two vectors is $\theta=90^\circ$. Hence, the work done by the force of gravity is zero.

Problem (4): A person pulls a crate using a force of $56\,{\rm N}$ which makes an angle of $25^\circ$ with the horizontal. The floor is frictionless. How much work does he do in pulling the crate over a horizontal distance of $200\,{\rm m}$?

Solution : The component of the external force parallel to the displacement does work on an object over a distance of $d$. In all work problems in physics, this force component parallel to the displacement is found by the formula $F_{\parallel}=F\cos \theta$. Thus, the work done by this force is computed as below \begin{align*} W&=F_{\parallel}d\\&=(F\cos\theta)d\\&=(56\cos 25^\circ)(200) \\&=10080\quad {\rm J}\end{align*} We could use the work formula from the beginning $W=Fd\cos\theta$ where $\theta$ is the angle between $F$ and $d$.

Problem (5): A worker pushes a cart with a force of $45\,{\rm N}$ directed at an angle of $32^\circ$ below the horizontal. The cart moves at a constant speed.  (a) Find the work done by the worker as the cart moves a straight distance of $50\,{\rm m}$.  (b) What is the net work done on the cart?

Solution :(a) All information to find the work done by the worker is given, so we have \begin{align*} W&=Fd\cos\theta\\&=(45)(50) \cos 32^\circ\\&=1912.5\quad {\rm J}\end{align*}  (b) ''net'' means "total". In all work problems in physics, there are two equivalent methods to find the net work. Identify all forces that are applied to the cart, find their resultant force, and then compute the work done by this net force over a specific distance.

Or compute all works done on the object across a distance individually, then sum them algebraically.

Usually, the second method is easier. We take this approach here.

The cart moves in a straight horizontal path. All forces apply on it are, the worker force $F$, the normal force $F_N$, and the force of gravity or its weight $F_g=mg$. The work done by normal and gravity forces in a horizontal displacement is always zero since the angle between these forces and the displacement is $90^\circ$. So, $W_N=W_g=0$. Hence, the net (total) work done on the object is $W_{total}=W_N+W_g+W_F=1912.5\,{\rm J}$

Problem (6): A $1200-{\rm kg}$ box is at rest on a rough floor. How much work is required to move it $5\,{\rm m}$ at a constant speed (a) along the floor against a $230\,{\rm N}$ friction force, (b) vertically?

Solution : In this problem, we want to displace a box $5\,{\rm m}$ horizontally and vertically. In the horizontal direction, there is also kinetic friction.

(a) At constant speed , means there is no acceleration in the course of displacement, so according to Newton's second law $\Sigma F=ma$, the net force on the box must be zero. To meet this condition, the external force $F_p$ applied by a person must cancel out the friction force $f_k$. So, $F_p=f_k=230\quad {\rm N}$ The force $F_p$ and displacement are both parallel, so their product get the work done by $F_p$ $W_p=F_p d=230\times 5=1150\quad {\rm J}$ (b) In the vertical path, two forces act on the box. One is the external lifting force, and the other is the force of gravity. Since the box is moving at constant speed vertically, there is no acceleration, and thus this lifting external force $F_p$ must be balanced with the weight of the box. $F_p=F_g=mg=(1200)(10)=12000\,{\rm J}$ Assume the box is moved vertically upward. In this case, the lifting force and displacement are parallel, so the angle between them is zero $\theta=0$, and the work done by this force is $W_p=F_p d\cos 0=12000\times 5=60\,{\rm kJ}$ On the other side, the weight force, or force of gravity $F_g=mg$ is always downward, so the angle between the box's weight and upward displacement is $180^\circ$. So, the work done by the weight of the box is \begin{align*}W_g&=F_g d\cos 180^\circ \\\\ &=(1200)(10)(5)(-1) \\\\ &=-60\,{\rm kJ}\end{align*} In such cases where the angle between $F$ and $d$ is $180^\circ$, they are called antiparallel.

Problem (7): A 40-kg crate is pushed using a force of 150 N at a distance of $6\,{\rm m}$ on a rough surface. The crate moves at a constant speed. Find (a) the work done by the external force on the crate. (b) The coefficient of kinetic friction between the crate and the floor?

Solution : (a) the crate is moved horizontally through a distance of $6\,{\rm m}$ by a force parallel to its displacement. So, the work done by this external force is $W_p=F_p d \cos\theta=(150)(6)\cos 0=900\,{\rm J}$ where subscript $p$ denotes the person or any external agent.

(b) According to the definition of the kinetic friction force formula, $f_k=\mu_k F_N$, to find the coefficient of kinetic friction $\mu_k$, we must have both the friction force and normal force $F_N$.

In the question, we are told that the crate moves at a constant speed, so there is no acceleration, and thus, the net force applied to it must be zero.

When the friction force, which opposes the motion, is equal to the external force $F_p$, then this condition is satisfied. So, $f_k=F_p=150\,{\rm N}$ On the other side, the crate is not lifted off the floor, so there is no motion vertically.

Balancing all forces applied vertically, the weight force and the normal force $F_N$, we can find the normal force $F_N$ as below \begin{gather*} F_N-F_g=0\\ F_N=F_g\\ \Rightarrow F_N=mg=40\times 10=400\quad {\rm N}\end{gather*} Therefore, the coefficient of kinetic friction is found to be $f_k=\frac{f_k}{F_N}=\frac{150}{400}=0.375$

Practice these questions to understand friction force Problems on the coefficient of friction

Problem (8): A 18-kg packing box is pulled at constant speed by a rope inclined at $20^\circ$. The box moves a distance of 20 m over a rough horizontal surface. Assume the coefficient of kinetic friction between the box and the surface to be $0.5$.  (a) Find the tension in the rope? (b) How much work is done by the rope on the box?

Solution : The aim of this problem is to find the work done by the tension in the rope. The magnitude of the tension in the rope is not given. So, we must first find it.

(a) We are told the box moves at a constant speed, so, as previously mentioned, the net force on the box must be zero to produce no acceleration. But what forces are acting horizontally on the box? The horizontal component of tension in the rope, $T_{\parallel}=T\cos\theta$, and the kinetic friction force $f_k$ in the opposite direction of motion are the forces acting on the box horizontally.

If these two forces are equal in magnitude but opposite in direction, then their resultant (net) becomes zero, and consequently, the box will move at a constant speed. \begin{align*} f_k&=T_{\parallel}\\\mu_k F_N&=T\cos\theta\quad (I) \end{align*} The forces in the vertical direction must also cancel each other since there is no motion vertically. As you can see in the figure, we have $F_N=T\sin\theta+F_g$ Substituting this into the relation (I), rearranging and solving for $T$, yields \begin{gather*} \mu_k (T\sin\theta+mg)=T\cos\theta \\\\ \Rightarrow T=\frac{\mu_k mg}{\cos\theta-\mu_k\sin\theta}\end{gather*} Substituting the numerical values into the above expression, we find the tension in the rope. $T=\frac{(0.5)(18)(10)}{\cos 20^\circ-(0.5) \sin20^\circ}=117\quad {\rm N}$  (b) The only force that causes the box to move some distance is the horizontal component of the tension in the rope, $T_{\parallel}=T\cos\theta$. So, the work done by the tension in the rope is \begin{align*} W&=T_{\parallel}d\\ &=(117)( \cos 20^\circ)(20) \\&=2199\quad {\rm J}\end{align*}

Problem (9): A table of mass 40 kg is accelerated from rest at a constant rate of $2\,{\rm m/s^2}$ for $4\,{\rm s}$ by a constant force. What is the net work done on the table?

Solution : This is a combination of a  kinematics problem and a physics work problem. Here, first, we must find the distance over which the box is displaced. The given information is: initial speed $v_0=0$, acceleration $2\,{\rm m/s^2}$, time taken $t=4\,{\rm s}$. Using this data, we can find the total displacement by applying the kinematics equation $\Delta x=\frac 12 at^2+v_0t$, \begin{align*} \Delta x&=\frac 12 at^2+v_0t\\\\&=\frac 12 (2)(4)^2+0(4)\\\\&=16\quad {\rm m}\end{align*} So, this constant force causes the table to move a distance of 16 meters across the surface. To find the work done, we need a force, as well. The force is mass times acceleration, $F=ma$, so we have $F=ma=40\times 2=80\,{\rm N}$ Now that we have both the force and displacement, the net work done on the table is the product of force along the displacement times the magnitude of displacement $W=80\times 16=128\quad {\rm J}$

## Work problems in a uniform circular motion

Problem (10): A 5-kg object is held at the end of a string and undergoes uniform circular motion around a circle of radius $5\,{\rm m}$. If the tangential speed of the object around the circle is $15\,{\rm m/s}$, how much work was done on the object by the centripetal force?

Solution : Here, an object moves around a circle, so we encounter a uniform circular motion problem .

In such motions around a curve or circle, that force in the radial direction exerting on the object is called the centripetal force.

On the other hand, a movement around a circle is tangent to the path at any instant of time. Thus, we conclude that in any uniform circular motion, a force is applied to the whirling object that is perpendicular to its motion at any moment of time.

So, the angle between the centripetal force and displacement at any instant is always zero, $\theta=0$. Using the work formula $W=Fd\cos\theta$, we find that the work done by the centripetal force is always zero.

This is another example of zero work in physics.

## Work problems on an incline

Problem (11): A $5-{\rm kg}$ box, initially at rest, slides $2.5\,{\rm m}$ down a ramp of angle $30^\circ$. The coefficient of friction between the box and the incline is $\mu_k=0.435$. Determine (a) the work done by the gravity force, (b) the work done by the frictional force, and (c) the work done by the normal force exerted by the surface.

Solution : This part is related to problems on inclined plane surfaces . The forces acting on a box on an inclined plane are shown in the figure. As you can see, the forces along the direction of motion are the parallel component of the weight $W_{\parallel}$, and the friction force $f_k$.

(a) In the figure, you realize that the angle between the object's weight (the same as the force of gravity) and downward displacement $d$ is zero, $\theta=30^\circ$.

So, the work done by the force of gravity on the box is found using work formula as below \begin{align*} W&=Fd\cos\theta\\&=(mg)(d) \cos 30^\circ\\&=(5\times 10)(2.5) \cos 30^\circ\\&=108.25\quad {\rm J}\end{align*}  (b) To find the work done by friction, we need to know its magnitude. From the kinetic friction force formula, $f_k=\mu_k F_N$, we must determine, first, the normal force acting on the box.

There is no motion in the direction perpendicular to the incline, so the resultant of forces acting in this direction must be zero. Equating the same direction forces, we will have $F_N=mg\sin\alpha=(5)(10) \sin 30^\circ=25\,{\rm N}$ Substituting this into the above equation for kinetic friction, we can find its magnitude as $f_k=\mu_k F_N=(0.435)(25)=10.875\,{\rm N}$ The friction force and the displacement of the box down the ramp are parallel, i.e., $\theta=0$. The work done by friction is \begin{align*} W_f&=f_kd\cos\theta \\\\ &=(10.875)(2.5) \cos 0\\\\ &=27.1875\,{\rm J}\end{align*}  (c) By definition, the normal force is the same contact force that is applied to the object from the surface perpendicularly. On the other hand, the object moves along the incline, so its displacement is perpendicular to the normal force, $\theta=90^\circ$. Hence, the work done by the normal force is zero. $W_N=F_N d\cos\theta=F_N d\cos 90^\circ=0$

Problem (12): We want to push a $950-{\rm kg}$ heavy object 650 m up along a $7^\circ$ incline at a constant speed. How much work do we do over this distance? Ignore friction.

Solution : When it comes to constant speed in all work problems in physics, you must remember that all forces in the same direction must be equal to the opposing forces. This condition ensures that there is no acceleration in the motion.

In this case, all forces acting on the object are: the pushing force along the incline upward $F_p$, and the parallel component of the force of gravity (weight) along the incline downward, $W_{\parallel}=mg\sin\alpha$. Thus, we can find the pushing force as \begin{align*}F_p&=mg\sin 7^\circ\\\\ &=(950)(10) \sin 7^\circ \\\\& =1159\quad {\rm N}\end{align*} In the question, we are told that the object is moving up the incline, so the angle between its displacement and upward pushing force is zero, $\theta=0$. Hence, the work done by the person to push the object along the incline upward is \begin{align*} W_p&=F_p d\cos\theta\\&=1159\times 650 \cos 0\\&=753350\quad{\rm J}\end{align*}

Problem (13): Consider an electron moving at a constant speed of $1.1\times 10^6\,\rm m/s$ in a straight line. How much energy is required to stop this electron? (Take the electron's mass, $m_e=9.11\times 10^{-31}\,\rm kg$.

Solution : In this problem on work, we cannot use the work formula directly, since none of the work variables, i.e., $F$, $d$, $\theta$, are given except the velocity. In these cases, we have a problem on the work-energy theorem .

According to this rule, the net work done over a distance by a constant force on an object of mass $m$ equals the change in its kinetic energy $W_{net}=\underbrace{\frac 12 mv_f^2-\frac 12 mv_i^2}_{\Delta K}$ Substituting the numerical values given in this problem, we get the required work to stop this fast-moving electron. \begin{align*}W_{net}&=\frac 12 m(v_f^2-v_i^2) \\\\ &=\frac 12 (9.11\times 10^{-31}) \left(0^2-(1.1\times 10^6)^2 \right) \\\\ &=-5.5\times 10^{-19}\,\rm J\end{align*} where we set the final velocity $v_f=0$ since the electron is to stop.

Problem (14): How much power is needed to lift a $25-\rm kg$ weight $1\,\rm m$ in $1\,\rm s$?

Solution : The power in physics is defined as the ratio of work done on an object to the time taken $P=\frac{W}{t}$. The SI unit of power is the watt ($W$).

In this problem, first, we must find the amount of work done in lifting the object as much as $1,\rm m$ vertically. The only force involved in this situation is the downward weight force. Thus, $W=(mg)h=(25\times 10)(1)=250\,{\rm J}$ This amount of work has been done in a time interval of $1\,\rm s$. Hence, the power is calculated as below $P=\frac{W}{t}=\frac{250}{1}=250\,\rm W$

Problem (15): A particle having charge $-3.6\,\rm nC$ is released from rest in a uniform electric field $E$ moves a distance of $5\,\rm cm$ through it. The electric potential difference between those two points is $\Delta V=+400\,\rm V$. What work was done by the electric force on the particle?

Solution : The work done by the electric force on a charged particle is calculated by $W_E=qEd$, where $E$ is the magnitude of the electric field and $d$ is the amount of distance traveled through $E$. But in this case, the electric field strength is not given, and we cannot use this formula.

We can see this as a problem on electric potential . Recall that the work done by the electric force on a charge to move it between two points with different potentials is given by $W=-q\Delta V$. Substituting the given numerical values into this, we will have \begin{align*} W&=-q\Delta V \\&=-(-3.6)(+400) \\&=\boxed{1440\,\rm J} \end{align*}

Here, we learned how to calculate the work done by a constant force in physics by solving a couple of example problems.

Overall, the work done by a constant force is the product of the horizontal component of the force times the displacement between the initial and final points.

In addition, power, a related quantity to work in physics, is also defined as the rate at which work is done.

Author : Dr. Ali Nemati Date Published : 9/20/2021

• Problems & Exercises
• Introduction to Science and the Realm of Physics, Physical Quantities, and Units
• 1.1 Physics: An Introduction

## 1.2 Physical Quantities and Units

1.3 accuracy, precision, and significant figures, 1.4 approximation.

• Section Summary
• Conceptual Questions
• Introduction to One-Dimensional Kinematics
• 2.1 Displacement
• 2.2 Vectors, Scalars, and Coordinate Systems
• 2.3 Time, Velocity, and Speed
• 2.4 Acceleration
• 2.5 Motion Equations for Constant Acceleration in One Dimension
• 2.6 Problem-Solving Basics for One-Dimensional Kinematics
• 2.7 Falling Objects
• 2.8 Graphical Analysis of One-Dimensional Motion
• Introduction to Two-Dimensional Kinematics
• 3.1 Kinematics in Two Dimensions: An Introduction
• 3.2 Vector Addition and Subtraction: Graphical Methods
• 3.3 Vector Addition and Subtraction: Analytical Methods
• 3.4 Projectile Motion
• Introduction to Dynamics: Newton’s Laws of Motion
• 4.1 Development of Force Concept
• 4.2 Newton’s First Law of Motion: Inertia
• 4.3 Newton’s Second Law of Motion: Concept of a System
• 4.4 Newton’s Third Law of Motion: Symmetry in Forces
• 4.5 Normal, Tension, and Other Examples of Forces
• 4.6 Problem-Solving Strategies
• 4.7 Further Applications of Newton’s Laws of Motion
• 4.8 Extended Topic: The Four Basic Forces—An Introduction
• Introduction: Further Applications of Newton’s Laws
• 5.1 Friction
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• Introduction to Uniform Circular Motion and Gravitation
• 6.1 Rotation Angle and Angular Velocity
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• 6.4 Fictitious Forces and Non-inertial Frames: The Coriolis Force
• 6.5 Newton’s Universal Law of Gravitation
• 6.6 Satellites and Kepler’s Laws: An Argument for Simplicity
• Introduction to Work, Energy, and Energy Resources
• 7.1 Work: The Scientific Definition
• 7.2 Kinetic Energy and the Work-Energy Theorem
• 7.3 Gravitational Potential Energy
• 7.4 Conservative Forces and Potential Energy
• 7.5 Nonconservative Forces
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• 7.8 Work, Energy, and Power in Humans
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• Introduction to Linear Momentum and Collisions
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• Introduction to Statics and Torque
• 9.1 The First Condition for Equilibrium
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• Introduction to Rotational Motion and Angular Momentum
• 10.1 Angular Acceleration
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• Introduction to Fluid Statics
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• 11.5 Pascal’s Principle
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• 11.9 Pressures in the Body
• Introduction to Fluid Dynamics and Its Biological and Medical Applications
• 12.1 Flow Rate and Its Relation to Velocity
• 12.2 Bernoulli’s Equation
• 12.3 The Most General Applications of Bernoulli’s Equation
• 12.4 Viscosity and Laminar Flow; Poiseuille’s Law
• 12.5 The Onset of Turbulence
• 12.6 Motion of an Object in a Viscous Fluid
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• Introduction to Temperature, Kinetic Theory, and the Gas Laws
• 13.1 Temperature
• 13.2 Thermal Expansion of Solids and Liquids
• 13.3 The Ideal Gas Law
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• Introduction to Heat and Heat Transfer Methods
• 14.2 Temperature Change and Heat Capacity
• 14.3 Phase Change and Latent Heat
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• Introduction to Thermodynamics
• 15.1 The First Law of Thermodynamics
• 15.2 The First Law of Thermodynamics and Some Simple Processes
• 15.3 Introduction to the Second Law of Thermodynamics: Heat Engines and Their Efficiency
• 15.4 Carnot’s Perfect Heat Engine: The Second Law of Thermodynamics Restated
• 15.5 Applications of Thermodynamics: Heat Pumps and Refrigerators
• 15.6 Entropy and the Second Law of Thermodynamics: Disorder and the Unavailability of Energy
• 15.7 Statistical Interpretation of Entropy and the Second Law of Thermodynamics: The Underlying Explanation
• Introduction to Oscillatory Motion and Waves
• 16.1 Hooke’s Law: Stress and Strain Revisited
• 16.2 Period and Frequency in Oscillations
• 16.3 Simple Harmonic Motion: A Special Periodic Motion
• 16.4 The Simple Pendulum
• 16.5 Energy and the Simple Harmonic Oscillator
• 16.6 Uniform Circular Motion and Simple Harmonic Motion
• 16.7 Damped Harmonic Motion
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• 16.10 Superposition and Interference
• 16.11 Energy in Waves: Intensity
• Introduction to the Physics of Hearing
• 17.2 Speed of Sound, Frequency, and Wavelength
• 17.3 Sound Intensity and Sound Level
• 17.4 Doppler Effect and Sonic Booms
• 17.5 Sound Interference and Resonance: Standing Waves in Air Columns
• 17.6 Hearing
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• Introduction to Electric Charge and Electric Field
• 18.1 Static Electricity and Charge: Conservation of Charge
• 18.2 Conductors and Insulators
• 18.3 Coulomb’s Law
• 18.4 Electric Field: Concept of a Field Revisited
• 18.5 Electric Field Lines: Multiple Charges
• 18.6 Electric Forces in Biology
• 18.7 Conductors and Electric Fields in Static Equilibrium
• 18.8 Applications of Electrostatics
• Introduction to Electric Potential and Electric Energy
• 19.1 Electric Potential Energy: Potential Difference
• 19.2 Electric Potential in a Uniform Electric Field
• 19.3 Electrical Potential Due to a Point Charge
• 19.4 Equipotential Lines
• 19.5 Capacitors and Dielectrics
• 19.6 Capacitors in Series and Parallel
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• Introduction to Electric Current, Resistance, and Ohm's Law
• 20.1 Current
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• 20.3 Resistance and Resistivity
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• 20.6 Electric Hazards and the Human Body
• 20.7 Nerve Conduction–Electrocardiograms
• Introduction to Circuits and DC Instruments
• 21.1 Resistors in Series and Parallel
• 21.2 Electromotive Force: Terminal Voltage
• 21.3 Kirchhoff’s Rules
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• 21.5 Null Measurements
• 21.6 DC Circuits Containing Resistors and Capacitors
• Introduction to Magnetism
• 22.1 Magnets
• 22.2 Ferromagnets and Electromagnets
• 22.3 Magnetic Fields and Magnetic Field Lines
• 22.4 Magnetic Field Strength: Force on a Moving Charge in a Magnetic Field
• 22.5 Force on a Moving Charge in a Magnetic Field: Examples and Applications
• 22.6 The Hall Effect
• 22.7 Magnetic Force on a Current-Carrying Conductor
• 22.8 Torque on a Current Loop: Motors and Meters
• 22.9 Magnetic Fields Produced by Currents: Ampere’s Law
• 22.10 Magnetic Force between Two Parallel Conductors
• 22.11 More Applications of Magnetism
• Introduction to Electromagnetic Induction, AC Circuits and Electrical Technologies
• 23.1 Induced Emf and Magnetic Flux
• 23.2 Faraday’s Law of Induction: Lenz’s Law
• 23.3 Motional Emf
• 23.4 Eddy Currents and Magnetic Damping
• 23.5 Electric Generators
• 23.6 Back Emf
• 23.7 Transformers
• 23.8 Electrical Safety: Systems and Devices
• 23.9 Inductance
• 23.10 RL Circuits
• 23.11 Reactance, Inductive and Capacitive
• 23.12 RLC Series AC Circuits
• Introduction to Electromagnetic Waves
• 24.1 Maxwell’s Equations: Electromagnetic Waves Predicted and Observed
• 24.2 Production of Electromagnetic Waves
• 24.3 The Electromagnetic Spectrum
• 24.4 Energy in Electromagnetic Waves
• Introduction to Geometric Optics
• 25.1 The Ray Aspect of Light
• 25.2 The Law of Reflection
• 25.3 The Law of Refraction
• 25.4 Total Internal Reflection
• 25.5 Dispersion: The Rainbow and Prisms
• 25.6 Image Formation by Lenses
• 25.7 Image Formation by Mirrors
• Introduction to Vision and Optical Instruments
• 26.1 Physics of the Eye
• 26.2 Vision Correction
• 26.3 Color and Color Vision
• 26.4 Microscopes
• 26.5 Telescopes
• 26.6 Aberrations
• Introduction to Wave Optics
• 27.1 The Wave Aspect of Light: Interference
• 27.2 Huygens's Principle: Diffraction
• 27.3 Young’s Double Slit Experiment
• 27.4 Multiple Slit Diffraction
• 27.5 Single Slit Diffraction
• 27.6 Limits of Resolution: The Rayleigh Criterion
• 27.7 Thin Film Interference
• 27.8 Polarization
• 27.9 *Extended Topic* Microscopy Enhanced by the Wave Characteristics of Light
• Introduction to Special Relativity
• 28.1 Einstein’s Postulates
• 28.2 Simultaneity And Time Dilation
• 28.3 Length Contraction
• 28.4 Relativistic Addition of Velocities
• 28.5 Relativistic Momentum
• 28.6 Relativistic Energy
• Introduction to Quantum Physics
• 29.1 Quantization of Energy
• 29.2 The Photoelectric Effect
• 29.3 Photon Energies and the Electromagnetic Spectrum
• 29.4 Photon Momentum
• 29.5 The Particle-Wave Duality
• 29.6 The Wave Nature of Matter
• 29.7 Probability: The Heisenberg Uncertainty Principle
• 29.8 The Particle-Wave Duality Reviewed
• Introduction to Atomic Physics
• 30.1 Discovery of the Atom
• 30.2 Discovery of the Parts of the Atom: Electrons and Nuclei
• 30.3 Bohr’s Theory of the Hydrogen Atom
• 30.4 X Rays: Atomic Origins and Applications
• 30.5 Applications of Atomic Excitations and De-Excitations
• 30.6 The Wave Nature of Matter Causes Quantization
• 30.7 Patterns in Spectra Reveal More Quantization
• 30.8 Quantum Numbers and Rules
• 30.9 The Pauli Exclusion Principle
• Introduction to Radioactivity and Nuclear Physics
• 31.2 Radiation Detection and Detectors
• 31.3 Substructure of the Nucleus
• 31.4 Nuclear Decay and Conservation Laws
• 31.5 Half-Life and Activity
• 31.6 Binding Energy
• 31.7 Tunneling
• Introduction to Applications of Nuclear Physics
• 32.1 Medical Imaging and Diagnostics
• 32.2 Biological Effects of Ionizing Radiation
• 32.3 Therapeutic Uses of Ionizing Radiation
• 32.5 Fusion
• 32.6 Fission
• 32.7 Nuclear Weapons
• Introduction to Particle Physics
• 33.1 The Yukawa Particle and the Heisenberg Uncertainty Principle Revisited
• 33.2 The Four Basic Forces
• 33.3 Accelerators Create Matter from Energy
• 33.4 Particles, Patterns, and Conservation Laws
• 33.5 Quarks: Is That All There Is?
• 33.6 GUTs: The Unification of Forces
• Introduction to Frontiers of Physics
• 34.1 Cosmology and Particle Physics
• 34.2 General Relativity and Quantum Gravity
• 34.3 Superstrings
• 34.4 Dark Matter and Closure
• 34.5 Complexity and Chaos
• 34.6 High-temperature Superconductors
• 34.7 Some Questions We Know to Ask
• A | Atomic Masses
• B | Selected Radioactive Isotopes
• C | Useful Information
• D | Glossary of Key Symbols and Notation

The speed limit on some interstate highways is roughly 100 km/h. (a) What is this in meters per second? (b) How many miles per hour is this?

A car is traveling at a speed of 33 m/s 33 m/s size 12{"33"" m/s"} {} . (a) What is its speed in kilometers per hour? (b) Is it exceeding the 90 km/h 90 km/h size 12{"90"" km/h"} {} speed limit?

Show that 1 . 0 m/s = 3 . 6 km/h 1 . 0 m/s = 3 . 6 km/h size 12{1 "." 0"m/s"=3 "." "6 km/h"} {} . Hint: Show the explicit steps involved in converting 1 . 0 m/s = 3 . 6 km/h. 1 . 0 m/s = 3 . 6 km/h. size 12{1 "." 0"m/s"=3 "." "6 km/h"} {}

American football is played on a 100-yd-long field, excluding the end zones. How long is the field in meters? (Assume that 1 meter equals 3.281 feet.)

Soccer fields vary in size. A large soccer field is 115 m long and 85 m wide. What are its dimensions in feet and inches? (Assume that 1 meter equals 3.281 feet.)

What is the height in meters of a person who is 6 ft 1.0 in. tall? (Assume that 1 meter equals 39.37 in.)

Mount Everest, at 29,028 feet, is the tallest mountain on the Earth. What is its height in kilometers? (Assume that 1 kilometer equals 3,281 feet.)

The speed of sound is measured to be 342 m/s 342 m/s size 12{"342"" m/s"} {} on a certain day. What is this in km/h?

Tectonic plates are large segments of the Earth’s crust that move slowly. Suppose that one such plate has an average speed of 4.0 cm/year. (a) What distance does it move in 1 s at this speed? (b) What is its speed in kilometers per million years?

(a) Refer to Table 1.3 to determine the average distance between the Earth and the Sun. Then calculate the average speed of the Earth in its orbit in kilometers per second. (b) What is this in meters per second?

Express your answers to problems in this section to the correct number of significant figures and proper units.

Suppose that your bathroom scale reads your mass as 65 kg with a 3% uncertainty. What is the uncertainty in your mass (in kilograms)?

A good-quality measuring tape can be off by 0.50 cm over a distance of 20 m. What is its percent uncertainty?

(a) A car speedometer has a 5.0 % 5.0 % size 12{5.0%} {} uncertainty. What is the range of possible speeds when it reads 90 km/h 90 km/h size 12{"90"" km/h"} {} ? (b) Convert this range to miles per hour. 1 km = 0.6214 mi 1 km = 0.6214 mi size 12{"1 km" "=" "0.6214 mi"} {}

An infant’s pulse rate is measured to be 130 ± 5 130 ± 5 size 12{"130" +- 5} {} beats/min. What is the percent uncertainty in this measurement?

(a) Suppose that a person has an average heart rate of 72.0 beats/min. How many beats does he or she have in 2.0 y? (b) In 2.00 y? (c) In 2.000 y?

A can contains 375 mL of soda. How much is left after 308 mL is removed?

State how many significant figures are proper in the results of the following calculations: (a) 106 . 7 98 . 2 / 46 . 210 1 . 01 106 . 7 98 . 2 / 46 . 210 1 . 01 size 12{ left ("106" "." 7 right ) left ("98" "." 2 right )/ left ("46" "." "210" right ) left (1 "." "01" right )} {} (b) 18 . 7 2 18 . 7 2 size 12{ left ("18" "." 7 right ) rSup { size 8{2} } } {} (c) 1 . 60 × 10 − 19 3712 1 . 60 × 10 − 19 3712 size 12{ left (1 "." "60" times "10" rSup { size 8{ - "19"} } right ) left ("3712" right )} {} .

(a) How many significant figures are in the numbers 99 and 100? (b) If the uncertainty in each number is 1, what is the percent uncertainty in each? (c) Which is a more meaningful way to express the accuracy of these two numbers, significant figures or percent uncertainties?

(a) If your speedometer has an uncertainty of 2 . 0 km/h 2 . 0 km/h size 12{2 "." 0" km/h"} {} at a speed of 90 km/h 90 km/h size 12{"90"" km/h"} {} , what is the percent uncertainty? (b) If it has the same percent uncertainty when it reads 60 km/h 60 km/h size 12{"60"" km/h"} {} , what is the range of speeds you could be going?

(a) A person’s blood pressure is measured to be 120 ± 2 mm Hg 120 ± 2 mm Hg size 12{"120" +- 2" mm Hg"} {} . What is its percent uncertainty? (b) Assuming the same percent uncertainty, what is the uncertainty in a blood pressure measurement of 80 mm Hg 80 mm Hg size 12{"80"" mm Hg"} {} ?

A person measures his or her heart rate by counting the number of beats in 30 s 30 s size 12{"30"" s"} {} . If 40 ± 1 40 ± 1 size 12{"40" +- 1} {} beats are counted in 30 . 0 ± 0 . 5 s 30 . 0 ± 0 . 5 s size 12{"30" "." 0 +- 0 "." 5" s"} {} , what is the heart rate and its uncertainty in beats per minute?

What is the area of a circle 3 . 102 cm 3 . 102 cm size 12{3 "." "102"" cm"} {} in diameter?

If a marathon runner averages 9.5 mi/h, how long does it take him or her to run a 26.22-mi marathon?

A marathon runner completes a 42 . 188 -km 42 . 188 -km size 12{"42" "." "188""-km"} {} course in 2 h 2 h size 12{2" h"} {} , 30 min, and 12 s 12 s size 12{"12"" s"} {} . There is an uncertainty of 25 m 25 m size 12{"25"" m"} {} in the distance traveled and an uncertainty of 1 s in the elapsed time. (a) Calculate the percent uncertainty in the distance. (b) Calculate the uncertainty in the elapsed time. (c) What is the average speed in meters per second? (d) What is the uncertainty in the average speed?

The sides of a small rectangular box are measured to be 1 . 80 ± 0 . 01 cm 1 . 80 ± 0 . 01 cm size 12{1 "." "80" +- 0 "." "01"" cm"} {} , {} 2 . 05 ± 0 . 02 cm, and 3 . 1 ± 0 . 1 cm 2 . 05 ± 0 . 02 cm, and 3 . 1 ± 0 . 1 cm size 12{2 "." "05" +- 0 "." "02"" cm, and 3" "." 1 +- 0 "." "1 cm"} {} long. Calculate its volume and uncertainty in cubic centimeters.

When non-metric units were used in the United Kingdom, a unit of mass called the pound-mass (lbm) was employed, where 1 lbm = 0 . 4539 kg 1 lbm = 0 . 4539 kg size 12{1" lbm"=0 "." "4539""kg"} {} . (a) If there is an uncertainty of 0 . 0001 kg 0 . 0001 kg size 12{0 "." "0001""kg"} {} in the pound-mass unit, what is its percent uncertainty? (b) Based on that percent uncertainty, what mass in pound-mass has an uncertainty of 1 kg when converted to kilograms?

The length and width of a rectangular room are measured to be 3 . 955 ± 0 . 005 m 3 . 955 ± 0 . 005 m size 12{3 "." "955" +- 0 "." "005"" m"} {} and 3 . 050 ± 0 . 005 m 3 . 050 ± 0 . 005 m size 12{3 "." "050" +- 0 "." "005"" m"} {} . Calculate the area of the room and its uncertainty in square meters.

A car engine moves a piston with a circular cross section of 7 . 500 ± 0 . 002 cm 7 . 500 ± 0 . 002 cm size 12{7 "." "500" +- 0 "." "002""cm"} {} diameter a distance of 3 . 250 ± 0 . 001 cm 3 . 250 ± 0 . 001 cm size 12{3 "." "250" +- 0 "." "001""cm"} {} to compress the gas in the cylinder. (a) By what amount is the gas decreased in volume in cubic centimeters? (b) Find the uncertainty in this volume.

How many heartbeats are there in a lifetime?

A generation is about one-third of a lifetime. Approximately how many generations have passed since the year 0 AD?

How many times longer than the mean life of an extremely unstable atomic nucleus is the lifetime of a human? (Hint: The lifetime of an unstable atomic nucleus is on the order of 10 − 22  s 10 − 22  s size 12{"10" rSup { size 8{ - "22"} } " s"} {} .)

Calculate the approximate number of atoms in a bacterium. Assume that the average mass of an atom in the bacterium is ten times the mass of a hydrogen atom. (Hint: The mass of a hydrogen atom is on the order of 10 − 27  kg 10 − 27  kg size 12{"10" rSup { size 8{ - "27"} } " kg"} {} and the mass of a bacterium is on the order of 10 − 15  kg. 10 − 15  kg. size 12{"10" rSup { size 8{ - "15"} } "kg"} {} )

Approximately how many atoms thick is a cell membrane, assuming all atoms there average about twice the size of a hydrogen atom?

(a) What fraction of Earth’s diameter is the greatest ocean depth? (b) The greatest mountain height?

(a) Calculate the number of cells in a hummingbird assuming the mass of an average cell is ten times the mass of a bacterium. (b) Making the same assumption, how many cells are there in a human?

Assuming one nerve impulse must end before another can begin, what is the maximum firing rate of a nerve in impulses per second?

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## High School Physics : Calculating Work

Study concepts, example questions & explanations for high school physics, all high school physics resources, example questions, example question #1 : calculating work.

In this case, there is only one force acting upon the object: the force due to gravity. Plug in our given information for the distance to solve for the work done by gravity.

Remember, since the object will be moving downward, the distance should be negative.

The work done is positive because the distance and the force act in the same direction.

## Example Question #2 : Calculating Work

Work is a force times a distance:

We know the distance that the book needs to travel, but we need to sovle for the lifting force required to move it.

There are two forces acting upon the book: the lifting force and gravity. Since the book is moving with a constant velocity, that means the net force will be zero. Mathemetically, that would look like this:

We can expand the right side of the equation using Newton's second law:

Use the given mass and value of gravity to solve for the lifting force.

Now that we have the force and the distance, we can solve for the work to lift the book.

This problem can also be solved using energy. Work is equal to the change in potential energy:

## Example Question #4 : Calculating Work

The relationship between work, force, and distance is:

We are given the force on the toy and the work done. Using these values, we can find the distance. Note that the mass is not relevant for this question.

## Example Question #6 : Calculating Work

The relationship between work, force and distance is:

We are given the value for the force and the distance that the toy travels. Using these values, we can find the work done by the cat. Note that the mass of the toy is not relevant for this calculation.

## Example Question #7 : Calculating Work

We are given the value for the work done by the cat and the distance that the toy travels. Using these values, we can find the force on the toy. Note that the mass of the toy is not relevant for this calculation.

## Example Question #8 : Calculating Work

The formula for work is:

Given the values for force and distance, we can calculate the work done.

Note that no work is done by the force of gravity or the weight of the box, since the vertical position does not change.

Work is the product of force times a distance:

We are given the work and the distance traveled, allowing us to solve for the force. The mass of the cabinet is not necessary information.

## Example Question #10 : Calculating Work

None of these

Use the data given to calculate the kinetic energy of the rocket at the two different velocities. Then find the amount of work done using the following equation:

Kinetic energy of the rocket at the two velocities:

The change in the kinetic energy at the two velocities:

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## AP®︎/College Physics 1

Welcome to ap/college physics 1, unit 1: kinematics and introduction to dynamics, unit 2: newton's laws, unit 3: circular motion and gravitation, unit 4: energy and momentum, unit 5: simple harmonic motion and rotational motion.

## Chapter: 11th Physics : UNIT 4 : Work, Energy and Power

Solved example problems for physics: work, energy and power, numerical problems.

1. Calculate the work done by a force of 30 N in lifting a load of 2kg to a height of 10m (g = 10ms -2 )

Force mg = 30 N ; height = 10 m

Work done to lift a load W = ?

W = F.S (or) mgh

W = 300 J

Ans:  300J

2. A ball with a velocity of 5 m s -1  impinges at angle of 60˚ with the vertical on a smooth horizontal plane. If the coefficient of restitution is 0.5, find the velocity and direction after the impact.

The impluse on the ball acts perpendicular to the smooth plane.

(i) The component of velocity of ball parallel to the surface.

(ii) For the component of velocity of ball perpendicular to the surface, apply law of restitution.

The component of velocity parallel to the surface will be changed.

v cos α = u cos 60°

v cos α = 5 × 1/2 = 5/2       ….(1)

According to law of restitution

v sin α = e u sin 60°

v sin α = 1/2 × 5 × √3/2 =  5 (√3/4)        ….(2)

Squaring and adding (1) and (2)

v 2 (sin 2 α + cos 2 α) =

v = 3.3 ms -1

Ans:  v = 0.3 m  s -1

3. A bob of mass m is attached to one end of the rod of negligible mass and length r, the other end of which is pivoted freely at a fixed center O as shown in the figure. What initial speed must be given to the object to reach the top of the circle? (Hint: Use law of conservation of energy). Is this speed less or greater than speed obtained in the section 4.2.9?

Ans: √ 4gr ms-1

The horizontal distance, draw the point of projection to the point where the ball returns to the same level

AC = OA-OC = r - rcosθ

minimum velocity = at v L = √[5 gr ]

v 1 2 = v L 2 - 2gr (1 - cos θ)

v 1 2 = 5 gr - 2gr (l-cosθ)

v 1 2 = 5gr-2gr(l – 1/2)

v 1 2 = 5gr - gr => v 1 2 = 4gr

v 1 = √[4 gr ] ms -1 .

4. Two different unknown masses A and B collide. A is initially at rest when B has a speed v. After collision B has a speed v/2 and moves at right angles to its original direction of motion. Find the direction in which A moves after collision.

Momentium is censerved in both × and y direction.

In x - direction

M B V B = O + M A V A ' cosɸ ...(l)

In y - direction

O = M B V B ' - M A V A 'sinɸ  ...(2)

(2)/(1) tanɸ = V B ’/V B = 1/2

ɸ = 26.6° (or) 26° 36' [1° = 60']

Ans:  θ   = 26° 33 ′

5. A bullet of mass 20 g strikes a pendulum of mass 5 kg. The centre of mass of pendulum rises a vertical distance of 10 cm. If the bullet gets embedded into the pendulum, calculate its initial speed.

Mass of the bullet m 1 = 20 g = 0.02 kg.

Mass of the pendulum m 2 = 5 kg

Centre of mass of pendulum rises to a height = h = 10 cm = 0.1 m

Speed of the bullet = u 1

Pendulum is at rest .:. u 2 = 0

Common velocity of the bullet and the pendulum after the bullet is embeded into the object = v

From II equation of motion

v = √[2gh] = √[2x9.8x0.1] = √[1.96] = 1.4 ms -1

Substitute the value of v in equation (1)

1.4 = 0.02 u 1 /5.02

u 1 = 5.02x1.4 / 0.02

u 1 = 351.4 ms -1

Ans:  v = 351.4m  s -1

## Conceptual Questions

1. A spring which in initially in un-stretched condition, is first stretched by a length x and again by a further length x. The work done in the first case W 1  is one third of the work done in second case W 2 . True or false?

2. Which is conserved in inelastic collision? Total energy (or) Kinetic energy?

Total energy is always conserved.

But K.E. is not conserved.

3. Is there any net work done by external forces on a car moving with a constant speed along a straight road?

If a car is moving at a constant speed, then external force will be zero.

Because a = [v - u] / t

For constant speed v = u , then a =0. ( a -acceleration)

F = ma   .'. F = zero. i.e.. no external force.

W = F.S. = 0. So net work done is zero.

4. A car starts from rest and moves on a surface with uniform acceleration.

Draw  the  graph  of  kinetic  energy versus displacement. What information you can get from that graph?

5. A charged particle moves towards another charged particle. Under what conditions the total momentum and the total energy of the system conserved?

(i) Both charged particles shall be dissimilar charge, (i.e. positive and negative)

(ii) After collision the charged particles; stick together permanent.

(iii) They should move with common velocity

## SOLVED EXAMPLE

2. A particle moves along X- axis from x=0 to x=8 under the influence of a force given by F= 3 x   2  - 4 x  + 5. Find the work done in the process.

3. A body of mass 10kg at rest is subjected to a force of 16N. Find the kinetic energy at the end of 10 s.

Mass m = 10 kg

Force F = 16 N

time t = 10 s

4. A body of mass 5kg is thrown up vertically with a kinetic energy of 1000 J. If acceleration due to gravity is 10 ms -2 , find the height at which the kinetic energy becomes half of the original value.

Mass m = 5kg

K.E          E = 1000J

g = 10 m s -2

5. Two bodies of mass 60 kg and 30 kg move in the same direction along straight line with velocity 40 cm s -1  and 30 cm   s -1   respectively suffer one dimensional elastic collision. Find their velocities after collision.

Mass m 1  = 60 kg

Mass m 2  = 30 kg

V 1  =  40cm s -1

V 2  =  30c m  s -1

Likewise,

6. A particle of mass 70 g moving at 50 cm s -1  is acted upon by a variable force as shown in the figure. What will be its speed once the force stops?

The area under the graph gives the impulse.

Impulse I = area of  ∆ OAE+ area of rectangle ABDE+ area of  ∆ DBC

But Impulse = 2 × initial momentum of the particle= 2 × m × u

Hence the particle will reverse its direction and move with its initial speed.

7. A particle strikes a horizontal frictionless floor with a speed u at an angle θ with the vertical and rebounds with the speed v at an angle Φ  with an vertical. The coefficient of restitution between the particle and floor is e. What is the magnitude of v?

Applying component of velocities,

The x - component of velocity is

The magnitude of y – component of velocity is not same, therefore, using coefficient of restitution,

8. A particle of mass m is fixed to one end of a light spring of force constant k and un-stretched length l. It is rotated with an angular velocity  ω   in horizontal circle. What will be the length increase in the spring?

Mass spring = m Force

constant = k

Un-stretched length = l

Angular velocity = ω

Let ‘x’ be the increase in the length of the spring.

The new length = (l+x) = r

When the spring is rotated in a horizontal circle,

Spring force = centripetal force.

9. A gun fires 8 bullets per second into a target X. If the mass of each bullet is 3 g and its speed 600 s -1 . Then, calculate the power delivered by the bullets.

Power = work done per second = total kinetic energy of 8 bullets per second

## Solved Example Problems for Work

Example 4.1.

A box is pulled with a force of 25 N to produce a displacement of 15 m. If the angle between the force and displacement is 30 o , find the work done by the force.

Force, F = 25 N

Displacement, dr = 15 m

Angle between F and dr, θ = 30 o

Work done ,  W  =  Fdr cos θ

## Solved Example Problems for Work done by a constant force

Example 4.2

An object of mass 2 kg falls from a height of 5 m to the ground. What is the work done by the gravitational force on the object? (Neglect air resistance; Take g = 10 m s -2 )

Work done by gravitational force is

The work done by the gravitational force on the object is positive.

Example 4.3

An object of mass  m = 1 kg is sliding from top to bottom in the frictionless inclined plane of inclination angle θ = 30 o  and the length of inclined plane is 10 m as shown in the figure. Calculate the work done by gravitational force and normal force on the object. Assume acceleration due to gravity, g = 10 m s -2

We calculated in the previous chapter that the acceleration experienced by the object in the inclined plane as g sin θ  .

According to Newton’s second law, the force acting on the mass along the inclined plane F  =  mg sin θ . Note that this force is constant throughout the motion of the mass.

The work done by the parallel component of gravitational force  (  mg sin θ )  is given by

## Example 4.4

If an object of mass 2 kg is thrown up from the ground reaches a height of 5 m and falls back to the Earth (neglect the air resistance). Calculate

a) The work done by gravity when the  object reaches 5 m height

b) The work done by gravity when the   object comes back to Earth

c) Total work done by gravity both in  upward and downward motion and mention the physical significance of the result.

When the object goes up, the displacement points in the upward direction whereas the gravitational force acting on the object points in downward direction. Therefore, the angle between gravitational force and displacement of the object is 180°.

a. The work done by gravitational force in the upward motion.

Given that ∆r =5 m and F mg

b. When the object falls back, both the gravitational force and displacement of the object are in the same direction. This implies that the angle between gravitational force and displacement of the object is 0°.

c. The total work done by gravity in the entire trip (upward and downward motion)

It implies that the gravity does not transfer any energy to the object. When the object is thrown upwards, the energy is transferred to the object by the external agency, which means that the object gains some energy. As soon as it comes back and hits the Earth, the energy gained by the object is transferred to the surface of the Earth (i.e., dissipated to the Earth).

## Example 4.5

A weight lifter lifts a mass of 250 kg with a force 5000 N to the height of 5 m.

a. What is the workdone by the weight lifter?

b. What is the workdone by the gravity?

c. What is the net workdone on the object?

a. When the weight lifter lifts the mass, force and displacement are in the same direction, which means that the angle between them θ = 0 0 . Therefore, the work done by the weight lifter,

b. When the weight lifter lifts the mass, the gravity acts downwards which means that the force and displacement are in opposite direction. Therefore, the angle between them θ = 180 0

c. The net workdone (or total work done) on the object

## Solved Example Problems for Work done by a variable force

Example 4.6.

A variable force  F   =  k x 2  acts on a particle which is initially at rest. Calculate the work done by the force during the displacement of the particle from  x   =  0 m to  x   =  4 m. (Assume the constant  k   = 1 N m -2 )

## Solved Example Problems for  Kinetic energy

Example 4.7.

Two objects of masses 2 kg and 4 kg are moving with the same momentum of 20 kg m s -1 .

a. Will they have same kinetic energy?

b. Will they have same speed?

a. The kinetic energy of the mass is given by

Note that  KE 1  ≠  KE 2  i.e., even though both are having the same momentum, the kinetic energy of both masses is not the same. The kinetic energy of the heavier object has lesser kinetic energy than smaller mass. It is because the kinetic energy is inversely proportional to the mass (KE   ∝  1/m) for a given momentum.

b. As the momentum,  p  =  mv , the two objects will not have same speed.

## Solved Example Problems for  Potential Energy

Example 4.8.

An object of mass 2 kg is taken to a height 5 m from the ground  g =  10ms -2  .

a. Calculate the potential energy stored in the object.

b. Where does this potential energy come from?

c. What external force must act to bring the mass to that height?

d. What is the net force that acts on the object while the object is taken to the height ‘h’?

a. The  potential energy  U   = m g h  = 2  ×  10  ×  5 = 100 J

Here the positive sign implies that the energy is stored on the mass.

b. This potential energy is transferred from external agency which applies the force on the mass.

d. From the definition of potential energy, the object must be moved at constant velocity. So the net force acting on the object is zero.

## Solved Example Problems for  Elastic Potential Energy

Example 4.9

Let the two springs A and B be such that k A >k B . On which spring will more work has to be done if they are stretched by the same force?

The work done on the springs are stored as potential energy in the springs.

k A >k B  implies that U B >U A  . Thus, more work is done on B than A.

## Example 4.10

A body of mass m is attached to the spring which is elongated to 25 cm by an applied force from its equilibrium position.

a. Calculate the potential energy stored in the spring-mass system?

b. What is the work done by the spring force in this elongation?

c. Suppose the spring is compressed to the same 25 cm, calculate the potential energy stored and also the work done by the spring force during compression. (The spring constant, k = 0.1 N m -1 ).

The spring constant, k = 0.1 N m -1

The displacement,  x  = 25 cm = 0.25 m

a. The potential energy stored in the spring is given by

Note that the potential energy is defined through the work done by the external agency. The positive sign in the potential energy implies that the energy is transferred from the agency to the object. But the work done by the restoring force in this case is negative since restoring force is in the opposite direction to the displacement direction.

c. During compression also the potential energy stored in the object is the same.

Work done by the restoring spring force during compression is given by

In the case of compression, the restoring spring force acts towards positive  x -axis and displacement is along negative  x  direction.

## Solved Example Problems for  Conservative and nonconservative forces

Example 4.11.

Compute the work done by the gravitational force for the following cases

(As the displacement is in two dimension; unit vectors and are used)

a. Since the motion is only vertical, horizontal displacement component d x  is zero. Hence, work done by the force along path 1 (of distance h).

Therefore, the total work done by the force along the path 2 is

Note that the work done by the conservative force is independent of the path.

## Example 4.12

Consider an object of mass 2 kg moved by an external force 20 N in a surface having coefficient of kinetic friction 0.9 to a distance 10 m. What is the work done by the external force and kinetic friction ? Comment on the result. (Assume g = 10 ms - 2 )

m  = 2 kg,  d  = 10 m,  F ext   = 20 N,    k   = 0.9.   When an object is in motion on the horizontal surface, it experiences two forces.

a. External force,  F ext   =  20 N

b. Kinetic friction,

f k  =μ k mg  = 0.9x(2)x10=18N.

The work done by the external force   W ext  = Fs = 20x20 =200J

The work done by the force of kinetic  friction  W k  = f k d  = (-18) x10=-180J  Here the negative sign implies that the  force of kinetic friction is  pposite to the  direction of displacement.

The total work done on the object

W total  = Wext + W k  = 200 J – 180 J = 20 J .

Since the friction is a non-conservative force, out of 200 J given by the external force, the 180 J is lost and it can not be recovered.

## Solved Example Problems for  Law of conservation of energy

Example 4.13.

An object of mass 1 kg is falling from the height  h  = 10 m. Calculate

a. The total energy of an object at  h   = 10 m

b. Potential energy of the object when it is at  h   =  4 m

c. Kinetic energy of the object when it is at  h   =  4 m

d. What will be the speed of the object when it hits the ground?

(Assume  g   = 10 m s -2 )

a. The gravitational force is a conservative force. So the total energy remains constant throughout the motion. At  h  = 10   m, the total energy E  is entirely   potential energy.

b. The potential energy of the object at  h  =   4   m is

c. Since the total energy is constant throughout the motion, the kinetic energy at  h   =  4 m must be  KE  =  E  -  U  =   100   -   40   =   60J

Alternatively, the kinetic energy could also be found from velocity of the object at 4 m. At the height 4 m, the object has fallen through a height of 6 m.

The velocity after falling 6 m is calculated from the equation of motion,

d. When the object is just about to hit the ground, the total energy is completely kinetic and the potential energy,  U   =  0.

## Example 4.14

A body of mass 100 kg is lifted to a height 10 m from the ground in two different ways as shown in the figure. What is the work done by the gravity in both the cases? Why is it easier to take the object through a ramp?

m = 100 kg, h = 10 m

Along path (1):

The minimum force  F 1  required to move the object to the height of 10 m should be equal to the gravitational force, F 1  mg = 100 x 10 = 1000 N

The distance moved along path (1) is,  = 10   m

The work done on the object along path (1) is

W = Fh = 1000 x 10 = 10,000 J

Along path (2):

In the case of the ramp, the minimum force  F 2   that we apply on the object to take it   up is not equal to  mg , it is rather equal to  mg  sin θ   . ( mg sin <  mg)  .

Here, angle θ = 30 o

Therefore, F 2   = mg sinθ = 100 × 10 ×  sin30 o  = 100 × 10 × 0.5 = 500N

Hence,  (mg sinθ < mg)

The path covered along the ramp is,

l = h/sin30 = 10/0.5 =20m

The work done on the object along path  (2) is, W = F2 l = 500 × 20 = 10,000 J

Since the gravitational force is a conservative force, the work done by gravity on the object is independent of the path taken.

In both the paths the work done by the gravitational force is 10,000 J

Along path (1): more force needs to be applied against gravity to cover lesser distance .

Along path (2): lesser force needs to be applied against the gravity to cover more distance.

As the force needs to be applied along the ramp is less, it is easier to move the object along the ramp.

## Example 4.15

An object of mass m is projected from the ground with initial speed v 0 .

Find the speed at height h.

Since the gravitational force is conservative; the total energy is conserved throughout the motion.

Final values of potential energy, kinetic energy and total energy are measured at the height  h .

By law of conservation of energy, the initial and final total energies are the same.

Note that in section (2.11.2) similar result is obtained using kinematic equation based on calculus method. However, calculation through energy conservation method is much easier than calculus method.

## Example 4.16

An object of mass 2 kg attached to a spring is moved to a distance  x   = 10 m from its equilibrium position. The spring constant  k  = 1   N m -1 and assume that the surface is   frictionless.

a. When the mass crosses the equilibrium position, what is the speed of the mass?

b. What is the force that acts on the object when the mass crosses the equilibrium position and extremum position  x  =  ±  10 m.

a. Since the spring force is a conservative force, the total energy is constant. At x  = 10   m, the total energy is purely   potential.

When the mass crosses the equilibrium position  x  = 0 , the potential energy

The entire energy is purely kinetic energy at this position.

b. Since the restoring spring force is F = - kx, when the object crosses the equilibrium position, it experiences no force. Note that at equilibrium position, the object moves very fast. When the object is at  x  = +10 m (elongation), the force F = - k  x

F = - (1) (10) = - 10 N. Here the negative sign implies that the force is towards equilibrium i.e., towards negative  x -axis and when the object is at  x  = - 10 (compression), it experiences a forces F = - (1) (- 10) = +10 N. Here the positive sign implies that the force points towards positive  x -axis.

The object comes to momentary rest at  x  =  ± 10 m  even though it experiences a maximum force at both these points.

## Solved Example Problems for  Motion in a vertical circle

Example 4.17.

Water in a bucket tied with rope is whirled around in a vertical circle of radius 0.5 m. Calculate the minimum velocity at the lowest point so that the water does not spill from it in the course of motion. (g = 10 ms -2 )

## Solved Example Problems for Unit of power

Example 4.18.

Calculate the energy consumed in electrical units when a 75 W fan is used for 8 hours daily for one month (30 days).

Power, P = 75 W

Time of usage, t = 8 hour × 30 days = 240 hours

Electrical energy consumed is the product of power and time of usage.

Electrical energy = power × time of usage = P × t

## Solved Example Problems for Relation between power and velocity

Example 4.19.

A vehicle of mass 1250 kg is driven with an acceleration 0.2 ms - 2  along a straight level road against an external resistive force 500 N . Calculate the power delivered by the vehicle’s engine if the velocity of the vehicle is 30 m  s - 1  .

The vehicle’s engine has to do work against resistive force and make vechile to move with an acceleration. Therefore, power delivered by the vehicle engine is

Solved Example Problems for collision

## Solved Example Problems for Elastic collisions in one dimension

Example 4.20.

A lighter particle moving with a speed of 10 m s -1  collides with an object of double its mass moving in the same direction with half its speed. Assume that the collision is a one dimensional elastic collision. What will be the speed of both particles after the collision?

Let the mass of the fi rst body be m which  moves with an initial velocity, u 1  = 10 m s -1 .

Therefore, the mass of second body is 2m and its initial velocity is u 2  = ½ u 1  = ½(10ms -1 )

Then, the fi nal velocities of the bodies can be calculated from the equation (4.53) and equation (4.54)

As the two speeds v 1 and v 2  are positive, they move in the same direction with the velocities, 3.33 m s −1  and 8.33 m s −1  respectively.

## Solved Example Problems for Perfect inelastic collision

Example 4.21.

A bullet of mass 50 g is fired from below into a suspended object of mass 450 g. The object rises through a height of 1.8 m with bullet remaining inside the object. Find the speed of the bullet. Take g = 10 ms -2 .

m 1  = 50 g = 0.05 kg; m 2  = 450 g = 0.45kg

The speed of the bullet is u 1 . The second body is at rest u 2  = 0 . Let the common velocity of the bullet and the object after the bullet is embedded into the object is v.

The combined velocity is the initial velocity for the vertical upward motion of the combined bullet and the object. From second equation of motion,

Substituting this in the above equation, the value of u 1  is

## Solved Example Problems for Coefficient of restitution (e)

Example 4.22.

Show that the ratio of velocities of equal masses in an inelastic collision when one of the masses is stationary is

v 1 /v 2  = 1-e/1+e

From the law of conservation of linear momentum,

Using the equation (2) for u 1  in (1), we get

Related Topics

Home O Level Work, Energy & Power Worked Examples for Work

• Worked Examples for Work
• Case Study 1: Energy Conversion for An Oscillating Ideal Pendulum
• Case Study 2: Energy Conversion for A Bouncing Ball
• Worked Examples for Energy

These worked examples will help to solidify your understanding towards the concept of work and work done .

## Worked Example 1: Pushing A Box

A boy pushes a box across a rough horizontal floor. he exerts 5 n to move it by 2 m. what is the work done by the boy.

\begin{aligned} \text{Work done by boy} &= f \times d \\ &= 5 \, \text{N} \times 2 \, \text{m} \\ &= 10 \, \text{J} \end{aligned}

## Worked Example 2: Pulling A Wagon

A boy pulls a 1 kg toy wagon along a smooth  (i.e. no friction) horizontal floor over a distance of 5 m. if the speed of the wagon increases at a constant rate of $2 \, \text{m s}^{-2}$, what is the work done by the boy.

The net resultant force on the toy wagon can be given by Newton’s Second Law : $f_{\text{resultant}} = ma$. Since the floor is smooth (no friction), the net resultant force on the toy wagon is just the force exerted by the boy.

\begin{aligned} \text{Net Resultant Force} &= \text{Force exerted by boy} \\ \text{Force exerted by boy} &= m \times a \\ &= 1 \, \text{kg} \times 2 \, \text{m s}^{-2} \\ &= 2 \, \text{N} \end{aligned}

The work done by the boy will just be:

\begin{aligned} \text{Work done by boy on wagon} &= \text{force} \times \text{dist. moved in direction of force} \\ &= 2 \, \text{N} \times 5 \, \text{m} \\ &= 10 \, \text{J} \end{aligned}

## Worked Example 3: Raising a Box

A 5 kg box is raised 50 m from its original position. what is the gain in gravitational potential energy assume gravitational field strength = $10 \, \text{n kg}^{-1}$. how much work is needed to raise the box to its new position.

Note: Assume that no energy is lost to air resistance.

\begin{aligned} E_{p} &= mgh \\ &= 5 \, \text{kg} \times 10 \, \text{N kg}^{-1} \times 50 \, \text{m} \\ &= 2500 \, \text{J} \end{aligned}

## Worked Example 4: Dropping A Ball

A ball with a mass of 500 g is dropped from a height of 10 m from the ground level (reference level)..

• What is it’s initial gravitational potential energy?
• Determine its velocity just before it hits the ground.

Taking the ground level to be the reference level,

(For gravitational potential energy , the relevant formula is:)

\begin{aligned} E_{p} &= mgh \\ &= \left( 0.5 \, \text{kg} \right) \left( 10 \, \text{N kg}^{-1} \right) \left( 10 \, \text{m} \right) \\ &= 50 \, \text{J} \end{aligned}

From the Principle of Conservation of Energy , the initial gravitational potential energy will be converted to its final kinetic energy. (I.e. Gain in kinetic energy = loss in gravitational potential energy)

For kinetic energy , the relevant formula is:

\begin{aligned} E_{k} &= \frac{1}{2} m v^{2} \\ 50 \, \text{J} &= \frac{1}{2} \left( 0.5 \, \text{kg} \right) v^{2} \\ v &= 14.1 \, \text{m s}^{-1} \end{aligned}

## Worked Example 5: Dropping A Ball With The Effects of Air Resistance

A ball with a mass of 500 g is dropped from a height of 10 m from the ground level (reference level). there is an energy loss of $10 \, \text{j}$ due to air resistance. determine its velocity just before it hits the ground..

You can obtain the gravitational potential energy of the ball from Worked Example 4, which is $50 \, \text{J}$.

By the Principle of Conservation of Energy , the initial gravitational potential energy will be converted to final kinetic energy and energy lost due to air resistance. Hence, we have:

\begin{aligned} E_{\text{initial}} &= E_{\text{final kinetic energy}} +\text{Energy Lost} \\ 50 \, \text{J} &= E_{\text{final kinetic energy}} + 10 \, \text{J} \\ E_{\text{final kinetic energy}} &= 50 \, \text{J}-10 \, \text{J} \\ &= 40 \, \text{J} \end{aligned}

Using the formula for kinetic energy , the velocity is:

\begin{aligned} E_{\text{final kinetic energy}} &= \frac{1}{2}mv^{2} \\ 40 \, \text{J} &= \frac{1}{2} \left( 0.5 \text{kg} \right) v^{2} \\ v &= 12.6 \, \text{m s}^{-1} \end{aligned}

## Worked Example 6: The Ball Rebounds

A ball with a mass of 500 g is thrown vertically downwards from a height of 10 m with a velocity of $5.0 \, \text{m s}^{-1}$. it hits the ground and bounces. it then rises to a maximum height of 8.0 m. determine the energy loss due to the bounce..

Assume that no energy is lost due to air resistance.

From the Principle of Conservation of Energy , the initial kinetic energy and initial gravitational potential energy is equals to the final kinetic energy + final gravitational potential energy + energy lost due to bounce. Let’s put this into an equation form:

\begin{aligned} E_{\text{k, initial}} + E_{\text{p, initial}} &= E_{\text{k, final}} + E_{\text{p, final}} + \text{Energy lost} \\ \frac{1}{2}mv^{2} + mgh_{\text{initial}} &= 0 + mgh_{\text{final}} + \text{Energy lost} \\ \frac{1}{2}\left( 0.5 \right) \left( 5^{2} \right) + \left( 0.5 \right) \left( 10 \right) \left( 10 \right) &= \left( 0.5 \right) \left( 10 \right) \left( 8 \right) + \text{Energy lost} \\ 56.25 \, \text{J} &= 40 \, \text{J} + \text{Energy lost} \\ \text{Energy lost} &= 16.25 \, \text{J} \\ \text{Energy lost} &= 16.3 \, \text{J} \, \left(3 \, \text{s.f.} \right) \end{aligned}

## Worked Example 7: Work Done By Car Engine

The engine of a car exerts a constant force of 10 kn. the car is able to accelerate constantly from rest to a speed of $30 \, \text{m s}^{-1}$ in 10 seconds. determine the work done by the engine of the car and the kinetic energy of the car at the end of the 10 seconds..

From what you have learnt in Kinematics ,

\begin{aligned} v &= u + at \\ a &= \frac{\left( 30-0 \right)}{10} \\ &= 3 \, \text{m s}^{-1} \end{aligned}

With the given engine force and calculated acceleration, we can find the mass of car by employing the use of Newton’s Second Law. ( Note: The engine force is the net resultant force acting on the car. Can you see why?)

\begin{aligned} F &= ma \\ m &= \frac{F}{a} \\ &= \frac{10000}{3} \\ &= 3330 \, \text{kg} \end{aligned}

The distance travelled by the car in 10 seconds is given by:

\begin{aligned} d &= \frac{1}{2} \left( v + u \right) \times t \\ &= 150 \, \text{m} \end{aligned}

The work done by the engine of the car will be:

\begin{aligned} W &= F \times d \\ &= 10000 \, \text{N} \times 150 \, \text{m} \\ &= 1 500 000 \, \text{J} \\ &= 1.5 \times 10^{6} \, \text{J} \end{aligned}

The kinetic energy of the car at the end of 10 seconds will be given by:

\begin{aligned} E_{k} &= \frac{1}{2} mv^{2} \\ &= \frac{1}{2} \left( 3330 \, \text{kg} \right) \left( 30 \, \text{m s}^{-1} \right)^{2} \\ &= 1.5 \times 10^{6} \end{aligned}

We notice that the work done by the engine of the car is the same as the kinetic energy of the car. This shows that all the work done by the engine of the car is converted into the kinetic energy of the car. (which is true if there is no dissipative forces acting on the car)

Back To Work, Energy And Power (O Level)

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• Published: 10 November 2023

## Seventy-five years of quantum electrodynamics

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On the 75th anniversary of quantum electrodynamics, Chad Orzel reflects on the parallel work that led to the synthesis of the theory.

Most physicists wouldn’t immediately think of 1948 as an important year in the history of quantum physics. However, 75 years ago this year saw the debut of quantum electrodynamics (QED), which completes the original project of quantum theory: explaining the structure of atoms and their interaction with light. The full version of QED fixes a problem that was uncovered soon after Paul Dirac’s successful combination of quantum mechanics and special relativity in 1930. Dirac’s equation applied in the simplest approximation agreed brilliantly with experiment, but, when physicists tried to incorporate effects like the electron interacting with its own electric field, they found nonsensical results: the predicted energy of even an isolated electron was infinite.

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Schwinger, J. On quantum-electrodynamics and the magnetic moment of the electron. Phys. Rev. 73 , 416 (1948).

Feynman, R. P. Space-time approach to quantum electrodynamics. Phys. Rev. 76 , 769 (1949).

Tomonaga, S.-I. & Oppenheimer, J. R. On infinite field reactions in quantum field theory. Phys. Rev. 74 , 224 (1948).

Dyson, F. J. The radiation theories of Tomonaga, Schwinger, and Feynman. Phys. Rev. 75 , 486 (1949).

Fan, X., Myers, T. G., Sukra, B. A. D. & Gabrielse, G. Measurement of the electron magnetic moment. Phys. Rev. Lett. 130 , 071801 (2023).

Guellati-Khelifa, S. Searching for new physics with the electron’s magnetic moment. Physics 16 , 22 (2023).

Aguillard, D. P. et al. Measurement of the positive muon Anomalous Magnetic Moment to 0.20 ppm. Preprint at https://doi.org/10.48550/arXiv.2308.06230 (2023).

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Orzel, C. Seventy-five years of quantum electrodynamics. Nat Rev Phys (2023). https://doi.org/10.1038/s42254-023-00664-2

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## Physical Chemistry Chemical Physics

A highly efficient, excellent water stability of mn 4+ activated nb-based oxyfluoride red fluorescent material preparation and performance analysis.

In this study, an efficient non-rare earth Mn 4+ -doped K 3 (NbOF 5 )(HF 2 ) red fluorescent material was synthesized by coprecipitation method. Replacing KF with K 2 CO 3 effectively solved the problem that KF was difficult to stir due to its strong water absorption. The sample was composed of the rods. The excitation spectra consisted of two strong excitation peaks at 366 nm and 468 nm. The emission spectra consisted of a series of narrow-band emission between 580 nm and 680 nm. Besides, the luminescence quantum efficiency (QE) reached 84.3% under the excitation of 468 nm. The fluorescent lifetime of K 3 (NbOF 5 )(HF 2 ):Mn 4+ was less than 4 ms, which can achieve fast response display in backlight display applications. The WLED was fabricated with K 3 (NbOF 5 )(HF 2 ):Mn 4+ and commercial YAG:Ce 3+ and the commercial InGaN blue chip. At 30 mA drive current, the WLED device achieved very excellent luminescence properties. The color temperature (CCT) was 3853 K, the color rendering index (Ra) was 90.1 and the luminous efficiency was 310.432 lm/W. Therefore, K 3 (NbOF 5 )(HF 2 ):Mn 4+ has a very broad prospect in WLED lighting and backlight display applications.

## Article information

R. Wang, H. Zhou, W. Shi, X. Yu, X. Mi, X. Liu and Y. Wang, Phys. Chem. Chem. Phys. , 2023, Accepted Manuscript , DOI: 10.1039/D3CP04631A

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## Work-Energy Practice Problems Video Tutorial

The Work-Energy Practice Problems Video Tutorial focuses on the use of work-energy concepts to solve problems. After providing a background and a short strategy, Mr. H steps through detailed solutions to six example problems involving work and energy. Learn how to use the concepts to solve for speed, height, and distance.  The video lesson answers the following question:

• How do you use the work-energy relationship to solve problems involving speed, height, and distance?

## View Video Tutorial

#### IMAGES

1. Learn AP Physics

2. Learn AP Physics

3. Bundle of Lessons

4. Physics.i.practice.problems.set.2.1

5. Physics Lesson

6. PHYSICS WORKSHEET:

#### VIDEO

1. 11 Physics

2. Physics Chapter 3 problem 2

3. THERMODYNAMICS PROBLEMS PHYSICS

4. Physics Example 4.7

5. Physics Chapter 3 problem 1

6. 7 Linear (1D) Motion Example Problems

1. Physics Work Problems for High Schools

Problem (1): A constant force of 1200 N is required to push a car along a straight line. A person displaces the car by 45 m. How much work is done by the person? Solution: If a constant force F F acts on an object over a distance of d d, and F F is parallel to d d, then the work done by force F F is the product of the force times distance.

2. Work, Energy, and Power Problem Sets

Problem Set WE1: Work 1 Use the work equation to calculate the work done, a force value, or a displacement value. Includes 8 problems. Problem Set WE2: Work 2 Use the work equation to calculate the work done, a force value, or a displacement value. Includes 6 problems. Problem Set WE3: Work and Power 1

3. Work, Energy, and Power Problem Sets

Problem 1: Renatta Gass is out with her friends. Misfortune occurs and Renatta and her friends find themselves getting a work out. They apply a cumulative force of 1080 N to push the car 218 m to the nearest fuel station. Determine the work done on the car. Audio Guided Solution Show Answer Problem 2:

4. Work example problems (video)

David goes through some example problems on the concept of work. By reviewing these, you'll have a better knowledge of how to calculate work done by individual forces on an object in motion.

5. 9.1 Work, Power, and the Work-Energy Theorem

W = f d. Some things that we typically consider to be work are not work in the scientific sense of the term. Let's consider a few examples. Think about why each of the following statements is true. Homework is not work. Lifting a rock upwards off the ground is work. Carrying a rock in a straight path across the lawn at a constant speed is not work.

6. PDF Physics Work Problems

a negative work implies that the motion is in the positive direction but the work done in the opposite direction is slowing the object down. For an incorrect statement, the opposite is true: the net work done to the particle must be zero or negative. A) Positive work is done so this is incorrect. B) Positive work is done so this is incorrect.

7. Calculating work done by a force (practice)

Course: Mechanics (Essentials) - Class 11th > Unit 7. Lesson 1: What does negative work even mean? Intro to work. Positive & negative work. The dot product. Calculating work done by a force. Work as area under curve. Calculating work from force vs. position graphs. Science >.

8. Work and Power: Problems

Problem : A 10 kg object experiences a horizontal force which causes it to accelerate at 5 m/s2, moving it a distance of 20 m, horizontally. How much work is done by the force? The magnitude of the force is given by F = ma = (10) (5) = 50 N.

9. What are energy and work? (article)

The definition of work, W , is below: W = F ⋅ Δ x. The work we need to do to burn the energy in the candy bar is E = 280 cal ⋅ 4184 J / cal = 1.17 MJ . Therefore, the distance, Δ x , we need to move the box through is: W = F ⋅ Δ x 1.17 MJ = ( 500 N) ⋅ Δ x 1.17 × 10 6 J 500 N = Δ x 2, 340 m = Δ x. Remember, however, that our ...

10. Work and energy

Physics library 19 units · 12 skills. Unit 1 One-dimensional motion. Unit 2 Two-dimensional motion. Unit 3 Forces and Newton's laws of motion. Unit 4 Centripetal force and gravitation. Unit 5 Work and energy. Unit 6 Impacts and linear momentum. Unit 7 Torque and angular momentum. Unit 8 Oscillations and mechanical waves.

11. Work

Hypertextbook Opus in profectus … frames work energy … Work discuss ion summary practice problems resources Problems practice Write something. Write something else. Write something different. Write something completely different. numerical An 11.3 g bullet leaves the muzzle of a 61 cm rifle with a horizontal velocity of 922 m/s. Determine…

12. Calculating Work in Physics: Example Problems

What is work in physics? We all do lots of work: homework, working on projects or working at a job. But in physics work has a very specific meaning. This vid...

13. Sample Problems

Problem 3.1. A toy train rolls along a straight, frictionless track, parallel to the x-axis. As it rolls, it experiences a force given by the equation: →F = F(x)[0.600 ˆi + 0.800ˆj] The function F(x) can be expressed by the graph below. The train starts at the position xo = 12m with a speed of 3.0m s, moving in the − x direction, and the ...

14. Ch. 1 Problems & Exercises

Introduction to Rotational Motion and Angular Momentum; 10.1 Angular Acceleration; 10.2 Kinematics of Rotational Motion; 10.3 Dynamics of Rotational Motion: Rotational Inertia; 10.4 Rotational Kinetic Energy: Work and Energy Revisited; 10.5 Angular Momentum and Its Conservation; 10.6 Collisions of Extended Bodies in Two Dimensions; 10.7 Gyroscopic Effects: Vector Aspects of Angular Momentum

15. Interactive Physics Example Problems

Two areas of modern physics are addressed through example problems on this page. Special Relativity problems ask you to relate the observations of two observers measuring the same thing. In Quantum Mechanics problems, you may look at wave or particle behavior of light and subatomic particles. As always, basic definitions problems are found with ...

16. Calculating Work

High School Physics Help » Calculating Work. . How much work was done on the box? , work equals force times distance. Remember, since the object will be moving downward, the distance should be negative. The work done is positive because the distance and the force act in the same direction. book falls off the top of a bookshelf.

17. Kinematic Equations: Sample Problems and Solutions

Kinematic Equations: Sample Problems and Solutions Kinematic equations relate the variables of motion to one another. Each equation contains four variables. The variables include acceleration (a), time (t), displacement (d), final velocity (vf), and initial velocity (vi).

18. Application and Practice Questions

Use the following diagram to answer questions #3 - #5. Neglect the effect of resistance forces. 3. As the object moves from point A to point D across the surface, the sum of its gravitational potential and kinetic energies ____. a. decreases, only. b. decreases and then increases. c. increases and then decreases.

19. AP®︎/College Physics 1

AP®︎/College Physics 1 5 units · 27 skills. Unit 1 Kinematics and introduction to dynamics. Unit 2 Newton's laws. Unit 3 Circular motion and gravitation. Unit 4 Energy and momentum. Unit 5 Simple harmonic motion and rotational motion. Course challenge. Test your knowledge of the skills in this course. Start Course challenge.

20. Solved Example Problems for Physics: Work, Energy and Power

Numerical Problems, Conceptual Questions, Solved Example Problems : Physics : Work, Energy and Power : Solved Example Problems for Physics: Work, Energy and Power Numerical Problems 1. Calculate the work done by a force of 30 N in lifting a load of 2kg to a height of 10m (g = 10ms-2) Answer: Given : Force mg = 30 N ; height = 10 m

21. Worked Examples for Work

Worked Example 7: Work Done By Car Engine. The engine of a car exerts a constant force of 10 kN. The car is able to accelerate constantly from rest to a speed of 30m s−1 30 m s − 1 in 10 seconds. Determine the work done by the engine of the car and the kinetic energy of the car at the end of the 10 seconds. Click to show/hide answer.

22. undefined Video Tutorial & Practice

Organic Chemistry. Start typing, then use the up and down arrows to select an option from the list. Learn undefined with free step-by-step video explanations and practice problems by experienced tutors.

23. Seventy-five years of quantum electrodynamics

Field theory ties together the whole Standard Model of particle physics, and extending the framework to incorporate gravity is the great open problem of theoretical physics. All of this work is ...

24. A highly efficient, excellent water stability of Mn4+ activated Nb

In this study, an efficient non-rare earth Mn 4+-doped K 3 (NbOF 5)(HF 2) red fluorescent material was synthesized by coprecipitation method.Replacing KF with K 2 CO 3 effectively solved the problem that KF was difficult to stir due to its strong water absorption. The sample was composed of the rods. The excitation spectra consisted of two strong excitation peaks at 366 nm and 468 nm.

25. Work, Energy and Power: Audio Guided Solution

Problem 1: Renatta Gass is out with her friends. Misfortune occurs and Renatta and her friends find themselves getting a work out. They apply a cumulative force of 1080 N to push the car 218 m to the nearest fuel station. Determine the work done on the car.

26. Physics Video Tutorial

The Work-Energy Practice Problems Video Tutorial focuses on the use of work-energy concepts to solve problems. After providing a background and a short strategy, Mr. H steps through detailed solutions to six example problems involving work and energy. Learn how to use the concepts to solve for speed, height, and distance.