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3.1Work

Work done by a force

We first consider the definition of work done is the product of the force in the direction of the displacement multiplied by the distance travelled:
W = Fs cos θ
where θ is the angle between the direction of the force, F, and the direction of the displacement, s by a constant force for motion in a straight line. By constant force we mean a force that is constant in magnitude as well as in direction. Figure 3.1 shows a block that is displaced along a straight line. The distance travelled by the body is s. The force makes an angle θ with the displacement.

Figure 3.1: A force moving its point of application performs work.

The force acts on the body all the time as it moves. The work done by the force is defined as

W=F s cos θ

But F cos θ is the component of the force in the direction of the displacement and so

KEY POINT

The work done by a force is the product of the force in the direction of the displacement times the distance travelled.

(Equivalently, since s cos θ is the distance travelled in the direction of the force, work may also be defined as the product of the force times the distance travelled in the direction of the force.)

The cosine here can be positive, negative or zero; thus work can be positive, negative or zero. We will see what that means shortly.

The unit of work is the joule. One joule is the work done by a force of 1 N when it moves a body a distance of 1 m in the direction of the force. 1 J = 1 N m.

Worked Example 3.1

A mass is being pulled along a level road by a rope attached to it in such a way that the rope makes an angle of 34° with the horizontal. The force in the rope is 24 N. Calculate the work done by this force in moving the mass a distance of 8.0 m along the level road.

Answer

We just have to apply the formula for work done:

W=F s cos θ

Substituting the values from the question,

W=24×8.0×cos 34°

W=160 J

Worked Example 3.2

A car with its engine turned off moves on a horizontal level road. A constant force of 620 N opposes the motion of the car. The car comes to rest after 84 m. Calculate the work done on the car by the opposing force.

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Answer

We again apply the formula for work done, but now we have to realise that θ = 180°. So

W=620×84×cos 180°

W=−52 kJ

Worked Example 3.3

You stand on roller skates facing a wall. You push against the wall and move away. Discuss whether the force exerted by the wall on you performed any work.

Answer

No work was done by the contact force because the point at which the force is applied has not moved. (So where did the energy you gained come from?)

Work done by a varying force

You will meet situations where the force is not constant in magnitude or direction and the path is not a straight line. Let us first discuss the case of a force varying in magnitude. Figure 3.2 shows how the magnitude of the force varies with distance travelled. Consider what happens when the force moves the body a very small distance Δs. Because Δs is so small we may assume that the force does not vary during this distance. The work done is then FΔs and is the area of the dark brown rectangle shown. For the total work we have to add the area of the many rectangles under the curve. The sum is the area under the curve.

Figure 3.2: The area under the graph of force against distance is the work done.

The work done by a force is the area under the graph that shows the variation of the magnitude of the force with distance travelled.

Worked Example 3.4

A force varies with distance travelled according to the graph in Figure 3.3. What is the work done in moving a distance of 4.0 m?

Answer

The work done is the area under the graph from d = 0 to d = 4 m. This is

W=2.0+10 2×4.0=24 J

Figure 3.3: The work done is the area under the graph. The area of a trapezoid is half the sum of the parallel sides multiplied by the perpendicular distance between them.

Work done in circular motion

We know that in circular motion there must be a force directed towards the centre of the circle. This is called the centripetal force.

Figure 3.4 shows the forces pointing towards the centre of the circular path. When we break the circular path into straight segments the angle between the force and the segment is always a right angle. This means that work done along each segment is zero because cos 90° = 0. So for circular motion the total work done by the centripetal force is zero.

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Figure 3.4: The work done by the centripetal force is zero.

CHECK YOURSELF 1

The work done by the centripetal force in a full revolution is zero because

athe displacement is zero

bthe force is at a right angle to the velocity

cthe centripetal force does not move its point of application

dforces do work only when they move the body along a straight line.

The work–kinetic energy relation

Imagine a net force F that acts on a particle of mass m. The force produces an acceleration a given by

a=F m

Let the initial speed of the particle be u. Because we have acceleration, the speed will change. Let the speed be v after travelling a distance s. We know from kinematics that

v 2=u 2+2 a s

Substituting for the acceleration, this becomes

v 2=u 2+2 F m s

We can rewrite this as

F s=1 2 m v 2−1 2 m u 2

We interpret this as follows: Fs is the work done on the particle by the net force. The quantity 1 2 × mass × speed2 is the energy the particle has due to its motion, called kinetic energy. For speed v, kinetic energy the energy possessed by a body that is moving:
E K=1 2 m v 2
where m is the mass of the body and v is its velocity E K is defined as

E k=1 2 m v 2

In our example, the initial kinetic energy of the particle is 1 2 m u 2 and the kinetic energy after travelling distance s is 1 2 m v 2. The result says that the work done has resulted in a transfer of energy that has changed the kinetic energy of the particle.

We can write this as

W net=Δ E K

where W net is the net work done and ΔE K is the change in kinetic energy. This is known as the work–kinetic energy relation the net work done on a system is equal to the change in kinetic energy of the system.

We can think of the work done as energy transferred. In this example, the work done has transferred energy to the particle by increasing its kinetic energy.

Worked example 3.5

A block of mass 2.5 kg slides on a rough horizontal surface. The initial speed of the block is 8.6 m s−1. It is brought to rest after travelling a distance of 16 m. Determine the magnitude of the frictional force.

Answer

We will use the work–kinetic energy relation, W net = ΔE K.

The only force doing work is the frictional force, f, which acts in the opposite direction to the motion.

W net=f×16×(−1)

The change in kinetic energy is

Δ E K=1 2 m v 2−1 2 m u 2=−92.45 J

So

−16 f=−92.45

f=5.8 N

The magnitude of the frictional force is 5.8 N.

(The angle between the force and the direction of motion is 180°, which is why we needed to multiply by cos 180°, which is –1.)

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Worked Example 3.6

You hold a ball of mass 0.25 kg in your hand and throw it so that it leaves your hand with a speed of 12 m s−1. Calculate the work done by your hand on the ball.

Answer

The question asks for work done, but here we do not know the forces that acted on the ball nor the distance by which we moved it before releasing it. But using ∆E K = W net, we find

W net=1 2 m v 2

W net=1 2×0.25×122=18 J

Worked Example 3.7

Suppose that in the previous example your hand moved a distance of 0.90 m in throwing the ball. Estimate the average net force that acted on the ball.

Answer

The work done was 18 J and so Fs = 18 J with s = 0.90 m. This gives F = 20 N.

Worked example 3.8

A mass m hangs from two strings attached to the ceiling such that they make the same angle with the vertical, as shown in Figure 3.5. The strings are shortened very slowly so that the mass is raised a distance h above its original position. Determine the work done by the tension in each string as the mass is raised.

Figure 3.5: For worked example 3.8.

Answer

The net work done is zero, either because the net force on the mass is zero or because the change in kinetic energy is zero. The work done by gravity is −mgh, and thus the work done by the two equal tension forces is +mgh. The work done by each is thus m g h 2.

Work done by gravity

We will now concentrate on the work done by a very special force, namely the weight of a body. Remember that weight is mass times acceleration of free fall and is directed vertically down. Thus, if a body is displaced horizontally, the work done by mg is zero. In this case the angle between the force and the direction of motion is 90° (Figure 3.6), so

W=m g s cos 90°=0

Figure 3.6: The force of gravity is normal to this horizontal displacement, so no work is being done.

Exam tip

When a body is displaced such that its final position is at the same vertical height as the original position, the work done by the weight is zero.

We are not implying that it is the weight that is forcing the body to move along the table. We are calculating the work done by a particular force (the weight) if the body (somehow) moves in a particular way.

If the body falls a vertical distance h, then the work done by the weight is +mgh. The force of gravity is parallel to the displacement, as in Figure 3.7a.

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If the body moves vertically upward to a height h from the initial position, then the work done by the weight is −mgh since now the angle between direction of force (vertically down) and displacement (vertically up) is 180°. The force of gravity is parallel to the displacement but opposite in direction, as in Figure 3.7b.

Suppose now that instead of just letting the body fall or throwing it upward, we use a rope to either lower it or raise it, at constant speed, by a height h (Figure 3.8). The work done by the weight is the same as before, so nothing changes. But we now ask about the work done by the force F that lowers or raises the body. Since F is equal and opposite to the weight, the work done by F is −mgh as the body is lowered and +mgh as it is being raised.

Figure 3.7: The force of gravity (green arrows) is parallel to the displacement in a and opposite in b.

Figure 3.8: Lowering and raising an object at constant speed using a rope.

In discussing work done it is always important to keep a clear picture of the force whose work we are calculating.

Gravitational potential energy

We just saw that when a force raises a body of mass m by a vertical distance h (at constant speed) the work done by that force is mgh.

For the earth–mass system we define the gravitational potential energy, E P the work done by a force in moving a body to a position above its initial position; for an earth–mass system,
E P = mgh
where m is the mass of the body, g is the gravitational field strength near the surface of the earth and h is the height above the earth’s surface; also the work done to bring two bodies to their present position from when they were infinitely far apart, to be the work done by the moving force in placing a body a height h above the surface of the earth: E P = mgh.

This implies that a body on the surface has zero potential energy. But as we will see (for example worked example 3.11 of the next section), we may call any horizontal surface to be the zero of potential energy and measure heights from that surface. This formula is approximate anyway and may only be used if the height h is very small compared to the radius of the earth.

Exam tip

Gravitational potential energy is the energy of a system due to its position and represents the work done by an external agent in bringing the system to that position.

Tension in a spring

Here we will meet another force to which we can associate a potential energy. It is the tension in a spring.

Consider a horizontal spring whose left end is attached to a vertical wall. If we apply a force F to the other end we will stretch the spring by some amount, x. We know from Chapter 2 that the force F and the extension x are directly proportional to each other, i.e., F = kx, a result known as Hooke’s law, Figure 3.9.

Since the force F and the extension x are directly proportional, the graph of force versus extension is a straight line through the origin and work done is the area under the curve (Figure 3.10).

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Figure 3.9: Stretching a spring requires work to be done.

Figure 3.10: The force F stretches the spring. The force is proportional to extension according to Hooke’s law.

To find the work done in extending the spring from its natural length (x = 0) to extension x, we need to calculate the area of the triangle of base x and height kx. Thus

area =1 2 k x×x area =1 2 k x 2

The work to extend a spring from its natural length by an amount x is thus

W=1 2 k x 2

The work done in extending the spring goes into elastic potential energy, E H the energy stored in a spring that has been stretched or compressed: E H=1 2 k x 2 where k is the spring constant and x is the amount by which the spring has been stretched or compressed stored in the spring. The elastic potential energy of a spring whose extension (or compression) is x is E H=1 2 k x 2. (The subscript H refers to Hooke).

Exam tip

Notice that in the data booklet the formula uses ∆x in place of our x. Notice also that the work done when extending a spring from an extension x 1 to an extension x 2 (so x 2 > x 1) is

W=1 2 k(x 2 2−x 1 2)

CHECK YOURSELF 2

A toy gun has a spring that is compressed by an amount e. The gun shoots a ball with speed v. What speed would the same ball have if the spring were compressed by an amount 2e?

Worked Example 3.9

A mass of 8.4 kg rests on top of a vertical spring whose base is attached to the floor. The spring compresses by 5.2 cm.

aCalculate the spring constant of the spring.

bDetermine the energy stored in the spring.

Answer

aThe mass is in equilibrium so mg = kx. So

k=m g x

k=8.4×9.8 5.2×10−2

k=1583≈1600 N m−1

bThe stored energy E H is

E H=1 2 k x 2

E H=1 2×1583×(5.2×10−2)2

E H=2.1 J

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TEST YOUR UNDERSTANDING

1A horizontal force of 24 N pulls a body a distance of 5.0 m along its direction. Calculate the work done by the force.

2A block slides along a rough table and is brought to rest after travelling a distance of 2.4 m. A force of 3.2 N opposes the motion. Calculate the work done by the opposing force.

3A block is pulled by a force making an angle of 20° to the horizontal, as shown in Figure 3.11. Find the work done by the pulling force when its point of application has moved 15 m.

Figure 3.11: For question 3.

4A block of mass m slides down an inclined plane a distance d, as shown in Figure 3.12. The incline makes an angle θ to the horizontal.

Figure 3.12: For question 4.

Determine the work done by the normal force N on the block.

5A ball of mass m is tied to a string and moves on a horizontal circle of radius R with constant speed v on a horizontal table.

aExplain carefully why the tension force does zero work during a full revolution.

bHow much work does it do in half a revolution?

6You are holding a weight in your palm. Discuss whether you are doing work.

7The graph (Figure 3.13) shows how the force F acting on a body varies with distance travelled x.

Figure 3.13: For question 7.

aCalculate the work done by the force after the body has been moved a distance of 5.0 m.

bIs the speed of the body increasing or decreasing?

8The velocity of an object moving along a straight line changes from 5.0 m s−1 to –5.0 m s−1 as the result of a force acting on the object. What is the work done by the force?

9A block of mass 2.0 kg and an initial speed of 5.4 m s−1 slides on a rough horizontal surface and is eventually brought to rest after travelling a distance of 4.0 m. Calculate the frictional force between the block and the surface.

10The graph (Figure 3.14) shows how the force F acting on a body varies with distance x travelled.

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Figure 3.14: For question 10.

When the body was at x = 0 its kinetic energy was 5.0 J. What is the kinetic energy at x = 5.0 m?

11A force F acts on a body of mass m = 2.0 kg initially at rest. The graph (Figure 3.15) shows how the force varies with distance travelled (along a straight line).

Figure 3.15: For question 11.

aFind the work done by this force.

bCalculate the final speed of the body.

12A body of mass 3.0 kg has kinetic energy 36 J. It is brought to rest after travelling a horizontal distance of 4.0 m. A frictional force acts in a direction opposite to the velocity. Determine the magnitude of the frictional force.

13Equal forces are applied to a 1 kg block and a 10 kg block, both of which are initially at rest. After moving a distance of 1 m, which body has the greatest kinetic energy, or is the energy the same for both?

14A body falls freely under the action of gravity. After falling 1 m the kinetic energy increases by 1 J. What will be the change in kinetic energy after falling another 1 m?

15A block of weight 150 N is tied to a rope. The block is lowered vertically by a distance of 12 m at constant speed. What is the work done by:

athe weight of the block

bthe tension in the rope.

16Figure 3.16 shows a block that is lowered by a vertical distance h along the curved red path.

Figure 3.16: For question 16.

By approximating the red path with the ‘staircase’ path, argue that the work done by the weight is still mgh and so show that the work done by the weight is independent of the path followed.

17You want to move a block to the top of an incline—from position A to position B. You can either pull it up the plane (Figure 3.17a) or raise it vertically (Figure 3.17b). In both cases, you use the same constant speed.

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Figure 3.17: For question 17.

In which case will the work done by gravity be the greatest—or is the work in both cases the same? What about the work done by you as you move the block?

18The extension of a spring of spring constant k = 250 N m−1 is 0.12 m. Calculate the elastic potential energy stored in the spring.

19The elastic potential energy stored in a spring of spring constant k = 380 N m−1 is 1.4 J. Calculate the extension of the spring.

20A spring of spring constant k = 200 N m−1 is slowly extended from an extension of 3.0 cm to an extension of 5.0 cm. Calculate the work done by the extending force.

21The graph (Figure 3.18) shows how the tension force F in a spring varies with the extension x.

Figure 3.18: For question 21.

Calculate the work done to stretch the spring from an extension of 5.0 cm to an extension of 15 cm.