Principles of Mechanics pp 53-72 | Cite as

# Work and Energy

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## Abstract

Energy is a very important concept that is heavily used in everyday life. Everything around us, including ourselves, needs energy to function. For example, electricity provides home appliances with the energy they require, food gives us energy to survive, and the sun provides earth with the energy needed for the existence of life! Experiments show that energy is a scalar quantity related to the state of an object. Energy may exist in various forms: mechanical, chemical, gravitational, electromagnetic, nuclear, and thermal. Furthermore, energy cannot be created or destroyed; it can only be transformed from one form to another. In other words, if energy were to be exchanged between objects inside a system, then the total amount of energy (the sum of all forms of energy) in the system will remain constant.

## 4.1 Introduction

Energy is a very important concept that is heavily used in everyday life. Everything around us, including ourselves, needs energy to function. For example, electricity provides home appliances with the energy they require, food gives us energy to survive, and the sun provides earth with the energy needed for the existence of life!

Experiments show that energy is a scalar quantity related to the state of an object. Energy may exist in various forms: mechanical, chemical, gravitational, electromagnetic, nuclear, and thermal. Furthermore, energy cannot be created or destroyed; it can only be transformed from one form to another. In other words, if energy were to be exchanged between objects inside a system, then the total amount of energy (the sum of all forms of energy) in the system will remain constant.

A transformation of energy occurs due to the action of a force known as work or due to heat exchange between objects (or between an object and its environment). If energy is transferred due to work then it may be defined as the capacity of doing work. This book is concerned with mechanical energy which involves kinetic energy (associated with the object’s motion) and potential energy (associated with the position of the object in space).

## 4.2 Work

Work may have many meanings. Sometimes, work is said to be done when a muscular activity is performed. Work may also refer to mental activity (mental work). In physics, the definition of work is different. Work is said to be done if a force is applied to an object while it is moving, i.e., if there is no resulting displacement, no work is done. Suppose that a person holds a heavy box for sometime and then starts to feel tired. The reason he/she feels tired is because chemical energy in his/her body is converted into internal microscopic motions of the muscles. Since the energy is not transferred to the box being carried (the box did not move), the work done on the box is equal to zero.

### 4.2.1 Work Done by a Constant Force

### 4.2.2 Work Done by Several Forces

### Example 4.1

A lady pulls an 80 kg block horizontally on a rough surface by a constant force of 400 \(\mathrm {N}\) that is at \(20^{\mathrm {o}}\) to the horizontal. If the block is pulled a distance of \(6\,\mathrm {m}\) and if the opposing force of friction has a magnitude of 118 \(\mathrm {N}:(\mathrm {a})\) determine the work done on the block by each of the applied force, the frictional force, the normal force, and the force of gravity; (b) find the total work done on the block; (c) determine if it is easier for the lady to pull the block at an angle larger than \(20^{\circ }\).

### Solution 4.1

(b) The total work done is

\(W_{tot}=W_{app}+W_{f}= (2255.3\,\mathrm {J})-(708\,\mathrm {J})=1547.3\,\mathrm {J}\)

The total work done can also be found by computing the net force acting on the block and calculating its work.

(c) For \((0\le \theta \le 90^{\mathrm {o}})\), If \(\theta _{2}>\theta _{1}\), then \(\cos \theta _{2}<\cos \theta _{1}\) and therefore \(W_{app2}< W_{app1}\), i.e., it is easier for the man to pull at an angle larger than \(20^{\mathrm {o}}\).

### Example 4.2

A delivery man wants to push a crate up a ramp of length *s*: (a) find the minimum work the man must do to lift the crate to the top of the ramp; (b) determine if a ramp with a steeper incline would be more difficult for the man to push the crate.

### Solution 4.2

(b) For angles between 0 and \(90^{\mathrm {o}}\), if \(\theta _{2}>\theta _{1}\), then \(\sin \theta _{2}>\sin \theta _{1}\). Hence \(W_{w2}\ge W_{w1}\), i.e., the more inclined the ramp is the more difficult it is to move the crate.

### 4.2.3 Work Done by a Varying Force

*n*of very small displacements where each is tangent to the path. For each displacement, the force can be approximated to be constant in both magnitude and direction. The total work done as the particle moves from \(\mathrm {P}\) to \(\mathrm {Q}\) is the sum of the individual amounts of work done along each displacement, that is

*x*,

*y*, and

*z*, and the curve can be determined by its equations that relates

*x*,

*y*, and

*z*to each other. The component form of the above equation is

*x*. Equation 4.1 is then reduced to

*F*(

*x*) is constant then we have

### Example 4.3

In Example 3.3, find the work done by the force in moving the particle during the time interval from \(t=0\) to \(t=1\,\mathrm {s}.\)

### Solution 4.3

### Example 4.4

A force acting on a particle is a function of position according to Fig. 4.6. Find the work done by this force as the particle moves from \(x_{i}=0\) to \(x_{f}=9\,\mathrm {m}.\)

### Solution 4.4

### Example 4.5

A ball that is suspended from a ceiling by a light rope is displaced a small distance to the position shown in Fig. 4.7. If it is released from rest at \(\mathrm {B}\), find the work done by the tension force and the force of gravity as the ball moves from \(\mathrm {B}\) to A.

### Solution 4.5

## 4.3 Kinetic Energy (KE) and the Work–Energy Theorem

*m*is moving with a speed

*v*, its KE is a scalar quantity defined as

The fact here is that it is incorrect to treat the man as a particle, since different parts of his body move in different ways as he pushes the bar. Therefore, the work–energy theorem does not hold. The man must be treated as a system of particles. In Chap. 6, we will see that the motion of a system of particles can be represented by the motion of its center of mass. The center of mass behaves as if all of the mass of the object (or system) is concentrated there and as if the net external force is applied there. In the case of the skateboarder, the center of mass of the system (man \(+\) skateboard) moves and the work–energy theorem can be applied to that point.

The work–energy theorem is an alternative method for describing motion without using Newton’s laws. It is especially useful in problems involving a varying force. Note that the work and the kinetic energy are not invariant quantities; they have different values when measured in different inertial frames of reference. However, from the principle of invariance, the equation \(W_{net}=\triangle K\) still holds for any inertial frame.

### Example 4.6

A 5 kg block resting on a surface is given an initial velocity of 5 \(\mathrm {m}/\mathrm {s}\). If the coefficient of kinetic friction of the surface is \(\mu _{k}=0.2\), find the distance the block would move before it stops.

### Solution 4.6

*s*is the displacement of the block.

### Example 4.7

A 10 kg block is pushed on a frictionless horizontal surface by a constant force of magnitude of 100 \(\mathrm {N}\) and that is at \(30^{\mathrm {o}}\) below the horizontal. If the block starts from rest, find its final speed after it has moved a distance of \(3\,\mathrm {m}\) using work–energy theorem.

### Solution 4.7

### 4.3.1 Work Done by a Spring Force

*x*of the block (or any other object) from its equilibrium position \((x=0)\). That is

*k*is a constant called the force or spring constant.

*k*measures the stiffness of the spring. The stiffer the spring the larger is

*k*. This equation is known as Hook’s law. The minus sign indicates that the spring force is always acting in a direction opposing the displacement. The work done by the spring force in moving the block from an initial position \(x_{i}\) to a final position \(x_{f}\) is:

*x*for the mass–spring system.

### Example 4.8

A 2 kg block is attached to a light spring of force constant 300 \(\mathrm {N}/\mathrm {m}\) on a horizontal smooth surface as shown in Fig. 4.11. If the system is initially at rest at the position of equilibrium and is then stretched a distance of 3 cm, find the work done by the spring on the block as it moves from \(x_{i}=0\) to \(x_{f}=3\) cm.

### Solution 4.8

### 4.3.2 Work Done by the Gravitational Force (Weight)

*m*is moving vertically upward or downward near the surface of the earth where \(\mathrm {g}\) is assumed to be constant (see Fig. 4.12), and if air resistance is neglected, then the only force that does work on the object is the gravitational force

*m*g. By taking the \(\mathrm {y}\)-axis along the line of motion (positive upwards) with \(y=0\) at the earth’s surface, the work done by the gravitational force is

### Example 4.9

A man lifts a 300 kg weight a distance of 2 \(\mathrm {m}\) above the ground. Find the work done by the force of gravity on the weight.

### Solution 4.9

### 4.3.3 Power

*W*on an object for a time interval \(\triangle t\), then the average power during that time is

## 4.4 Conservative and Nonconservative Forces

- 1.
The net work done by a conservative force on a particle moving from one point to another is independent of the path taken by the particle;

- 2.
The net work done by a conservative force in moving a particle through any closed path is equal to zero.

A force not meeting these conditions is a nonconservative force. As mentioned in Sect. 1.10.2, property 2 of a conservative force can be obtained from property 1 (if \(\mathbf {A}\) is a vector field and the line integral of \(\mathbf {A}\) between any two points is independent of path, then \(\displaystyle \oint _{C}\mathbf {A}\cdot \mathbf {r}=0\)). That is, these two properties are equivalent. Examples of conservative forces in mechanics are the gravitational and spring forces. To show this let us go back to Sects. 4.3.1 and 4.3.2, where the work done by the gravitational force or the spring force was calculated. We have seen that the work done in each case depends only on the initial and final positions of the object. Therefore, the work done by any of these forces is independent of the path joining the initial and final positions. Furthermore, if \((x_{i}=x_{f})\) in the case of the spring or \((y_{i}=y_{f})\) in the case of the gravitational force the net work done is zero. Hence, these forces are conservative.

### 4.4.1 Potential Energy

For a system consisting of two or more objects, the potential energy *U* of the system is the energy associated with the configuration of the system. That is, the potential energy is the energy associated with the position of objects in the system relative to each other. If the configuration of the system is changed, then the potential energy of the system also changes. Such energy is defined only in terms of a conservative force because if such a force acts on a system then it can transform the kinetic energy of any object in the system into potential energy of the system and vice versa. The potential energy means that the system has potential to do work.

*U*) is chosen. This is because if \(U_{i}\) is changed \(U_{f}\) will be also changed but \(\triangle U\) will remain constant.

### Example 4.10

A force acting on a particle is given by \(\mathbf {F}=-k\mathbf {r}\). Determine: (a) whether or not the force is conservative; (b) the potential energy associated with the force if it is conservative.

### Solution 4.10

### Example 4.11

If a force acting on a particle is given by \(\mathbf {F}=ay\mathbf {j}\), where *a* is a positive constant: (a) find the work done in moving the particle along the closed path shown in Fig. 4.16; (b) determine if the force is conservative.

### Solution 4.11

### Example 4.12

Find the force acting on a particle if the potential energy associated with it is \(U=5y^{2}-3z.\)

### Solution 4.12

#### 4.4.1.1 The Gravitational Potential Energy

#### 4.4.1.2 The Elastic Potential Energy

## 4.5 Conservation of Mechanical Energy

### 4.5.1 Changes of the Mechanical Energy of a System due to External Nonconservative Forces

### 4.5.2 Friction

*s*is the displacement of the object in the system. The reason behind not being able to calculate the work done by friction is that at a microscopic level the frictional force is not a single force that acts at one point. Rather, it is a combination of forces acting at different points in the object. However, the loss in kinetic energy of the object can be calculated as shown below: Consider a block sliding on a rough surface. Let’s choose the block only to be our system. From the equation of motion, we have

### 4.5.3 Changes in Mechanical Energy due to Internal Nonconservative Forces

*energy cannot be created or destroyed it can only be transformed from one form to another and the total energy of an isolated system is conserved*(

*constant*). This statement is known as the law of conservation of energy The law of conservation of energy is also valid in relativity and quantum mechanics.

### 4.5.4 Changes in Mechanical Energy due to All Forces

### Example 4.13

A 0.2 kg apple falls from a tree at a distance of 3 \(\mathrm {m}\) above the ground. Find: (a) the velocity of the apple at an altitude of 2 \(\mathrm {m}\) and at the instance just before it hits the ground; (b) the altitude of the apple when its velocity is 4 \(\mathrm {m}/\mathrm {s}.\)

### Solution 4.13

*h*is its initial altitude. That gives

### Example 4.14

A roller coaster of mass 500 kg starts from rest at point \(\mathrm {A}\), and rolls down the track as shown in Fig. 4.18. Ignoring friction, determine: (a) the roller coaster speed at \(\mathrm {B}\) and \(\mathrm {C}\); (b) the work done by gravity as the rollercoaster moves from A to B.

### Solution 4.14

### Example 4.15

A block of mass 5 kg is released from rest at the top of a \(45^{\mathrm {o}}\) incline that is 0.5 \(\mathrm {m}\) long as shown in Fig. 4.19. It then slides on a horizontal surface that is 0.7 \(\mathrm {m}\) long and goes up again on a second ramp that is at \(30^{\mathrm {o}}\) to the horizontal. If the coefficient of kinetic friction between the block and all three surfaces is 0.2, find the maximum distance that the block would move up the second ramp?

### Solution 4.15

### Example 4.16

Two masses \(m_{1}=5\) kg and \(m_{2}=9\) kg are connected by a light rope that passes over a massless frictionless pulley as in Fig. 4.20. If the system is released from rest when \(m_{2}\) is at 0. \(5\,\mathrm {m}\) above the ground, use the principle of conservation of energy to determine the speed with which \(m_{2}\) will hit the ground.

### Solution 4.16

### Example 4.17

A 0.25 kg ball is attached to alight string of length \(L=0.5\,\mathrm {m}\) as in Fig. 4.21. Find (a) the tension in the string at \(\mathrm {B}(\theta =10^{\circ })\) if the ball is given an initial velocity \(v_{a}=0.5\,\mathrm {m}/\mathrm {s}\) at its lowest position; (b) the velocity of the ball at A if the ball is released from rest at B.

### Solution 4.17

### Example 4.18

A 3 kg block compresses a spring of negligible mass a distance of 0.1 \(\mathrm {m}\) from its equilibrium position as in Fig. 4.22. If the surface is frictionless and the force constant of the spring is 200 \(\mathrm {N}/\mathrm {m}\), and the block is free to move, find: (a) the speed of the block just as it leaves the spring; (b) the maximum height that the block will reach; (c) suppose that a part of the horizontal track is rough with a length of 0.05 m, find the coefficient of kinetic friction if the block reaches a maximum height of 0.014 \(\mathrm {m}\).

### Solution 4.18

### Example 4.19

A small stone of mass 0.1 kg is released from rest inside a large hemispherical bowl of radius \(R=0.2\,\mathrm {m}\). It then slides along the surface as in Fig. 4.23. (a) Find the speed of the stone at point \(\mathrm {B}\) and \(\mathrm {C}\); (b) If the surface of the bowl is not frictionless, how much energy is dissipated by friction as the stone moves from A to \(\mathrm {B}\) if the speed at \(\mathrm {B}\) is 1.7 \(\mathrm {m}/\mathrm {s}\)?

### Solution 4.19

\(\displaystyle \triangle K_{ext}=\frac{1}{2}mv_{b}^{2}-mgR=(0.1\,\mathrm {k}\mathrm {g})\left( \frac{1}{2}(1.7\,\mathrm {m}/\mathrm {s})^{2}-(9.8\,\mathrm {m}/\mathrm {s}^{2})(0.2\,\mathrm {m})\right) =-0.05\,\mathrm {J}\)

### Example 4.20

A skier starts at the top of a frictionless incline as in Fig. 4.24. Find the velocity with which he will leave the second incline.

### Solution 4.20

### Example 4.21

A 0.4 kg stone is released from rest at point A where \(h_{A}=2\,\mathrm {m}\) (see Fig. 4.25). It then slides without friction along the track shown where \(R=0.5\,\mathrm {m}\). Determine: (a) the speed of the stone at \(\mathrm {B}\); (b) the normal force exerted on the stone at \(\mathrm {B}\); (c) the magnitude of the total acceleration of the stone at \(\mathrm {C}\); (d) the minimum height in which the stone must be released such that it does not fall off the track.

### Solution 4.21

### 4.5.5 Power

### 4.5.6 Energy Diagrams

*x*) is shown in Fig. 4.26. At any point

*F*(

*x*) is given by

*U*versus

*x*curve, the kinetic energy at any point can be found by subtracting the value of

*U*(at that certain point) from

*E*.

### 4.5.7 Turning Points

A turning point is a point in which the particle changes its direction of motion. The points \(x_{1}\), \(x_{3}\), \(x_{5}\) and \(x_{7}\) are all turning points.

### 4.5.8 Equilibrium Points

Equilibrium points occur in general when \(\nabla U=0\). In the case of one dimensional motion it occurs when \(dU(x)/dx=0\), i.e. when \(F(x)=0.\)

### 4.5.9 Positions of Stable Equilibrium

If at an equilibrium point \(d^{2}U(x)/dx^{2}>0\), then *U*(*x*) is a minimum at that point. The point is then said to be a position of stable equilibrium, i.e., any minimum on the *U*(*x*) curve is a position of stable equilibrium. Another method to find the position of stable equilibrium is to find the sign of *F*(*x*) at each side of the point. As an example, consider the point \(x_{2}\).

This point is a position of stable equilibrium since if the particle is displaced slightly to the right of \(x_{2}\) then *dU*(*x*) / *dx* is positive which leads to *F*(*x*) being negative and the particle will accelerate back towards \(x_{2}\). On the other hand, if the particle is displaced slightly to the left of \(x_{2}\), then *dU*(*x*) / *dx* is negative and thus *F*(*x*) is positive and the particle will also accelerates back to \(x_{2}\). Therefore, because *F*(*x*) tends to restore the particle back to that position when the particle is displaced in either direction, it is called a position of stable equilibrium. \(x_{6}\) is also a position of stable equilibrium.

### 4.5.10 Positions of Unstable Equilibrium

*U*(

*x*) is maximum at that point, and the point is called a position of unstable equilibrium. In Fig. 4.26, \(x_{4}\) is a position of unstable equilibrium since if the particle is slightly displaced to the right of \(x_{4}, F(x)\) is positive and the particle will accelerate away from \(x_{4}\). If the particle is displaced to the left of \(x_{4}, F(x)\) is negative and the particle will also accelerate away from that position. Therefore, because

*F*(

*x*) tends to repel the particle away from that position, it is called a position of unstable equilibrium. In general this force tends to move the particle towards the minimum value of

*U*(

*x*). Figure 4.27 shows the potential energy of a mass–spring system as a function of

*x*.

### 4.5.11 Positions of Neutral Equilibrium

*U*(

*x*) is constant and \(F(x)=0\) is called a position of neutral equilibrium. \(x_{8}\) is a position of neutral equilibrium. If the particle is slightly displaced to the right or left of \(x_{8}\), no restoring or repelling forces will act on the particle and it will remain stationary The position of the particle as a function of time can be obtained from

*x*we get the position as a function of time.

### Example 4.22

Figure 4.28 shows the potential energy of a particle as a function of its displacement. Find: (a) the values of *x* where the particle is in stable or unstable equilibrium; (b) the direction of the force acting on the particle at 0.5 \(\mathrm {m}.\)

### Solution 4.22

(a) We have \(x=1\,\mathrm {m}\) and \(x=4\,\mathrm {m}\) are positions of stable equilibrium, \(x=3\,\mathrm {m}\) is a position of unstable equilibrium.

(b) At 0.5 \(\mathrm {m}, dU(x)/dx\) is negative and hence *F*(*x*) is positive which means that the particle will accelerate in the positive \(\mathrm {x}\)-direction.

### Example 4.23

Consider a block attached to a light spring and released from rest at \(x=A.\) Find the position of the block as a function of time using energy methods.

### Solution 4.23

**Problems**

- 1.
A force acting on a particle varies with position as in Fig. 4.29. Find the work done by the force as the particle moves from \(x=0\) to \(x=8\,\mathrm {m}.\)

- 2.
A force \(\mathbf {F}=(3\mathbf {i}+\mathbf {j}-5\mathbf {k})\,\mathrm {N}\) acts on a particle that undergoes a displacement \(\mathbf {r}=(-2\mathbf {i}+3\mathbf {j}-\mathbf {k})\,\mathrm {m}\). Find the work done by the force on the particle.

- 3.
A 5 \(\mathrm {k}\mathrm {g}\) block is pulled from rest on a rough surface by a constant force of 10 \(\mathrm {N}\) that is at \(30^{\mathrm {o}}\) to the horizontal. If the coefficient of kinetic friction between the block and the surface is 0.15, find the final speed of the block as it moves through a displacement of \(2\,\mathrm {m}\) using the work–energy theorem.

- 4.
Calculate the work done against gravity in moving a 30 kg box through a height of 6 \(\mathrm {m}.\)

- 5.
A 1600 kg car accelerates from rest at a rate of 1 \(\mathrm {m}/\mathrm {s}^{2}\). Find the average power delivered to the car during the first 5 \(\mathrm {s}.\)

- 6.
Determine whether or not the force \(\mathbf {F}=-m\omega ^{2}(x\mathbf {i}+y\mathbf {j})\) is conservative, where \(\omega \) is constant and

*m*is the mass of the particle. If the force is conservative determine the potential energy associated with it. - 7.
A 5 \(\mathrm {k}\mathrm {g}\) block slides down an inclined plane of angle \(50^{\mathrm {o}}\) (see Fig. 4.30). Using energy methods, find the speed of the block just as it reaches the bottom if the coefficient of kinetic friction is \(\mu _{k}=0.2.\)

- 8.
A block of mass of 2 kg is pressed against a light spring of force constant 400 \(\mathrm {N}/\mathrm {m}\) (see Fig. 4.31). If the compression of the spring is 10 cm, find the maximum height the block will reach when it is released.

- 9.
A force acting on a particle is given by \(\mathbf {F}=-\beta y^{2}\mathbf {j}\). Find the work done in moving the particle along the path shown in Fig. 4.32.

- 10.
Two blocks are connected by a light rope that passes over a massless frictionless pulley (see Fig. 4.33). If the system is released from rest, find the total kinetic energy of the blocks when the 5 kg block descends a distance of 0.5 \(\mathrm {m}\) assuming that the surface is frictionless.

- 11.
A particle of mass 1.5 kg moves along the \(\mathrm {x}\)-axis where its potential energy varies as in Fig. 4.34. Plot the force \(F_{x}(x)\) versus

*x*from \(x=0\) to \(x=8\,\mathrm {m}.\) - 12.
A block of mass

*m*rests on a hemispherical mound of ice as shown in Fig. 4.35. If it is given a very small push and start sliding, find the height of the point in which the block will lose contact with the mound. - 13.
A 3 kg block hangs from a spring as in Fig. 4.36. If the spring stretches a distance of 10 cm, find (a) the force constant of the spring (b) the work done in expanding the spring a distance of 5 cm without accelerating it.

- 14.
In Fig. 4.37, determine the Turning points and the positions of stable and unstable equilibrium.

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