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A.3

Work, energy and power

Theme A · Space, time and motion · SL and HL

🎯What you need to be able to do

  • Calculate the work done by a force, including when the force is at an angle to the displacement.
  • Read work off a force–distance graph as an area.
  • Use kinetic energy, gravitational potential energy and elastic potential energy.
  • Apply conservation of energy to problems where more than one store changes.
  • Calculate power, including \( P = Fv \), and calculate efficiency.
  • Interpret and draw Sankey diagrams.

📚The physics

Work is a transfer, not a thing. A body does not contain work; work is what happens when a force moves its point of application.

\[ W = Fs\cos\theta \]

where \(\theta\) is the angle between the force and the displacement. Three consequences fall straight out of that cosine and they are worth more marks than the formula itself. A force perpendicular to the motion does no work — which is why the tension in a string does no work on a ball whirled in a circle, and why the normal force does no work on a sliding block. A force opposing the motion does negative work, as friction does. And if nothing moves, no work is done however hard you push.

When the force varies, the formula stops working and you use the graph instead: the work done is the area under the force–distance graph. For a spring obeying Hooke’s law that area is a triangle, which is exactly where the \( \tfrac{1}{2}kx^{2} \) comes from.

The three stores you need.

\( E_k = \tfrac{1}{2}mv^{2} \)
\( \Delta E_p = mg\Delta h \)
\( E_p = \tfrac{1}{2}kx^{2} \)

Note the square on the kinetic energy, so doubling the speed quadruples the energy, which is the whole argument behind speed limits. The gravitational form applies near the Earth’s surface, where \(\Delta h\) is the vertical height change and the path taken is irrelevant. In the elastic form, \(x\) is the extension from natural length, not the total length.

A useful identity: since \( W = \Delta E_k \), the resultant force multiplied by distance gives the change in kinetic energy. That is often a one-line route to an answer that would take three suvat steps.

Conservation of energy is the most powerful tool in this theme, because it skips the middle of a problem entirely. A ball rolling down a curved track has a horrible force analysis and a trivial energy analysis: \(mgh\) at the top becomes \( \tfrac{1}{2}mv^{2} \) at the bottom, so \( v = \sqrt{2gh} \), and the shape of the track never enters the calculation. Add friction and the same statement still works — the missing energy has gone to internal energy, and you can calculate exactly how much.

Power is the rate of energy transfer, \( P = W/t \). The version that catches people out is \( P = Fv \), which follows from \( W = Fs \) divided by \(t\). It is the form to reach for whenever a question mentions a vehicle moving at constant speed: at constant speed the driving force equals the resistive force, so the engine power equals resistance × speed.

Efficiency is useful output divided by total input, as a fraction or a percentage, and it is never greater than 1. A Sankey diagram shows the same information as a picture: the width of each arrow is proportional to the energy per second, the arrows always sum to the input, and the wasted branches point downwards. If your arrows do not add up, the diagram is wrong.

✏️Worked example

A 65 kg cyclist freewheels from rest down a hill 24 m high and reaches 18 m s\(^{-1}\) at the bottom. Take \( g = 9.81 \) m s\(^{-2}\).

(a) How much energy was dissipated by resistive forces? The store lost is gravitational: \( mg\Delta h = 65 \times 9.81 \times 24 = 15300 \) J. The store gained is kinetic: \( \tfrac{1}{2} \times 65 \times 18^{2} = 10500 \) J. The difference, 4.8 kJ, has gone to internal energy in the brakes, tyres and air.

(b) What was the average resistive force, if the slope is 150 m long? Work done against resistance is force × distance along the slope, so \( F = 4800/150 = 32 \) N.

(c) What is the efficiency of the descent? Useful output 10500 J, total input 15300 J, so efficiency \( = 0.69 \), or 69%.

The trap. In part (b) the distance is the 150 m travelled along the slope, not the 24 m of height. The height belongs to the gravitational calculation; the resistive force acts along the path.

🔭See it happen

PhET, Energy Skate Park. Turn the bar chart on and watch kinetic and potential trade places while the total stays flat. Then add friction and watch the thermal bar grow to fill exactly the gap. That picture is what part (a) above is doing algebraically.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. A 200 N force is applied at 25° to the horizontal to drag a crate 12 m horizontally. Find the work done by that force.
\( W = Fs\cos\theta = 200 \times 12 \times \cos 25^\circ = 2.2 \times 10^{3} \) J. Only the component along the displacement does work.
2. A 1200 kg car travelling at 20 m s\(^{-1}\) brakes to rest in 40 m. Find the average braking force.
The kinetic energy to remove is \( \tfrac{1}{2}mv^{2} = \tfrac{1}{2}(1200)(20^{2}) = 2.4 \times 10^{5} \) J. Since \( W = Fs = \Delta E_k \), \( F = 2.4 \times 10^{5}/40 = 6.0 \times 10^{3} \) N. Using energy skips the need for any suvat step.
3. A spring of stiffness 250 N m\(^{-1}\) is extended by 12 cm. Find the elastic potential energy stored.
\( E_p = \tfrac{1}{2}kx^{2} = \tfrac{1}{2} \times 250 \times 0.12^{2} = 1.8 \) J. Note the extension in metres, not centimetres.
4. A pendulum bob is released from a point 0.35 m above its lowest position. Find its speed at the lowest point, ignoring air resistance.
\( mgh = \tfrac{1}{2}mv^{2} \), so \( v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 0.35} = 2.6 \) m s\(^{-1}\). The mass cancels, and the shape of the swing is irrelevant.
5. A car travels at a constant 25 m s\(^{-1}\) against a total resistive force of 600 N. Find the power developed by the engine.
At constant speed the driving force equals the resistance, so \( P = Fv = 600 \times 25 = 1.5 \times 10^{4} \) W, or 15 kW.
6. An electric motor draws 500 W and raises a 12 kg load at a steady 1.5 m s\(^{-1}\). Find its efficiency.
Useful output power \( = mgv = 12 \times 9.81 \times 1.5 = 177 \) W. Efficiency \( = 177/500 = 0.35 \), or 35%. The missing 65% goes to heat in the windings, friction and sound.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • The Physics Classroom — Work, Energy and Power
  • The Physics Hypertextbook — energy