3.3 Work, energy and power — Learn

Prerequisites: the equations of motion from Section 3.1; resolving a vector into components from Section 2.2; the idea of energy transfer from Chapter 1.

By the end of this section you can: calculate the work done by a constant force, including a force not along the line of motion; use the kinetic energy and gravitational potential energy equations; apply conservation of energy; and calculate power from work and time, and from force and velocity.

1. Work done by a constant force

Work is done when a force moves its point of application. For a constant force $F$ that acts at an angle $\theta$ to the displacement $s$ ,

W = Fs\cos\theta

Work is a scalar, measured in joules (J). One joule is one newton metre.

1\,\text{J} = 1\,\text{N}\,\text{m}

The $s\cos\theta$ part is the component of the displacement along the force. Only the component of the force along the direction of motion does work. When the force is along the displacement, $\theta = 0$ and $\cos 0 = 1$ , so $W = Fs$ . When the force is perpendicular to the displacement, $\theta = 90^\circ$ and $\cos 90^\circ = 0$ , so the force does no work.

Work can be negative. A friction force acts opposite to the motion, so $\theta = 180^\circ$ , $\cos 180^\circ = -1$ , and the work done by friction is negative. Negative work removes energy from the object and transfers it to the surroundings, usually as thermal energy.

Worked example 1. A force of 40 N pushes a box 5.0 m along a floor. Find the work done.

W = Fs = 40 \times 5.0 = 200\,\text{J}

Worked example 2. A rope pulls a sled with a force of 250 N at 30° to the horizontal. The sled moves 12 m along the ground. Find the work done by the rope.

\begin{aligned} W &= Fs\cos\theta \\ &= 250 \times 12 \times \cos 30^\circ \\ &= 3000 \times 0.866 = 2600\,\text{J} \end{aligned}

Only the horizontal component of the pull, $250\cos 30^\circ$ , does work, because the sled moves horizontally.

2. Work transfers energy

Work is a way of transferring energy. The work–energy theorem says that the total work done on an object equals its change in kinetic energy.

For a constant force along the line of motion, $W = Fs = (ma)s$ . Using $v^2 = u^2 + 2as$ gives $as = \dfrac{v^2 - u^2}{2}$ , so

W = m\left(\frac{v^2 - u^2}{2}\right) = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mu^2

The work done equals the gain in kinetic energy,

E_k = \tfrac{1}{2}mv^2

When a force raises an object through a vertical height $\Delta h$ , the work done against gravity is stored as gravitational potential energy:

\Delta E_p = mg\Delta h

Near the Earth's surface, $g$ is constant, so a change in height gives a change in potential energy. Only changes in potential energy matter; the zero is chosen for convenience.

Worked example 3. A car of mass 1500 kg speeds up from $10\,\text{m}\,\text{s}^{-1}$ to $20\,\text{m}\,\text{s}^{-1}$ . Find the work done by the resultant force.

\begin{aligned} E_k(\text{initial}) &= \tfrac{1}{2} \times 1500 \times 10^2 = 75000\,\text{J} \\ E_k(\text{final}) &= \tfrac{1}{2} \times 1500 \times 20^2 = 300000\,\text{J} \\ W &= 300000 - 75000 = 225000\,\text{J} = 225\,\text{kJ} \end{aligned}

Worked example 4. A book of mass 2.0 kg is lifted through 1.5 m. Find the gravitational potential energy gained. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .

\Delta E_p = mg\Delta h = 2.0 \times 9.81 \times 1.5 = 29\,\text{J}

3. Conservation of energy

Energy is transferred, not created or destroyed. The total energy of a closed system stays the same. In a frictionless system the gravitational potential energy lost by a falling object equals the kinetic energy gained.

If friction or drag is present, some energy is transferred to the thermal store of the surroundings. The mechanical energy then falls, but the total energy, including the thermal energy, is still conserved.

Worked example 5. A ball is dropped from a height of 5.0 m. Find its speed just before it lands, if air resistance is negligible.

The gravitational potential energy lost equals the kinetic energy gained.

\begin{aligned} mgh &= \tfrac{1}{2}mv^2 \\ v &= \sqrt{2gh} = \sqrt{2 \times 9.81 \times 5.0} \\ &= 9.9\,\text{m}\,\text{s}^{-1} \end{aligned}

The mass cancels, so the landing speed is the same for a heavy ball and a light one.

4. Power

Power is the rate of transfer of energy.

P = \frac{W}{t}

Power is measured in watts (W). One watt is one joule per second. For an object moving at a constant velocity $v$ under a driving force $F$ , the power is

P = Fv

because the force moves through a distance $Fs$ in time $t$ .

Worked example 6. A pump raises 200 kg of water through 15 m in 30 s. Find the useful power of the pump. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .

\begin{aligned} W &= mg\Delta h = 200 \times 9.81 \times 15 = 29430\,\text{J} \\ P &= \frac{W}{t} = \frac{29430}{30} = 981\,\text{W} \end{aligned}

Worked example 7. A car travels at a constant $25\,\text{m}\,\text{s}^{-1}$ with a driving force of 2000 N. Find the useful power developed.

P = Fv = 2000 \times 25 = 50000\,\text{W} = 50\,\text{kW}

At a constant speed the driving force balances the resistive forces, so the useful power is the rate at which the engine works against them.

Watch out. A force that is perpendicular to the motion does no work, even when it is large. A person carrying a heavy box along a level floor does no work on the box against gravity, because the upward force and the horizontal displacement are at right angles. Check that the force has a component along the displacement before you use $W = Fs$ .

5. Try it: work from a force–displacement graph

A force–displacement graph turns work into an area. Put the component of the force along the displacement on the vertical axis and the displacement on the horizontal axis. The area under the graph is the work done. For a constant force the graph is a horizontal line, so the area is a rectangle with height $F\cos\theta$ and width $s$ .

Predict first. Do not move the sliders yet. Write down your answers.

A force of $100\,\text{N}$ pulls at $60^\circ$ to the displacement. Is the work done more or less than $100 \times s$ ?
As the angle rises from $0^\circ$ to $90^\circ$ , does the work done rise, fall, or stay the same?
At $90^\circ$ the force is as large as before. What is the work done, and why?

Test. Set the force to $250\,\text{N}$ , the distance to $12\,\text{m}$ and the angle to $30^\circ$ , and check the readout against Worked example 2. Then keep the force and the distance fixed and move the angle through $0^\circ$ , $60^\circ$ and $90^\circ$ . Watch the height of the rectangle and its area.

Explain. Why does the height of the rectangle fall as the angle grows, when the applied force does not change? Write two or three sentences. Use the idea of the component of the force along the displacement.

Progress check

Return to the chapter problem. The car and riders have a mass of 500 kg and are released from rest 45 m above the bottom of the loop.

Find the gravitational potential energy lost by the car as it descends 45 m.
Assuming no energy is transferred to thermal stores, find the kinetic energy and the speed of the car at the bottom.
The top of the loop is 24 m above the bottom. Find the speed of the car at the top of the loop.

If question 1 goes wrong, use $\Delta E_p = mg\Delta h$ . If question 2 goes wrong, set the potential energy lost equal to the kinetic energy gained, and rearrange for $v$ . If question 3 goes wrong, first find the potential energy at the top of the loop, then subtract it from the energy at the bottom.

You should now be able to

calculate work done by a constant force, including a force not along the line of motion;
explain why a perpendicular force does no work;
use the work–energy theorem and the kinetic energy equation;
calculate gravitational potential energy changes;
apply conservation of energy, including energy transferred to thermal stores;
calculate power from work and time, and from force and velocity;
explain why the mass cancels in a free-fall energy calculation.