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M 5

Energy, work and power

Mechanics · Paper 4

🎯What you need to be able to do

  • Calculate the work done by a constant force, including one at an angle to the motion.
  • Calculate kinetic energy and gravitational potential energy.
  • Apply conservation of mechanical energy where no resistance acts.
  • Use the work–energy principle when friction or a driving force is present.
  • Calculate power, and use \( P = Fv \) for a vehicle.
  • Find a maximum speed from a power output and a resistance.

📚The mathematics

Work done

\[ \text{work done} = Fd\cos\theta \]

where \( \theta \) is the angle between the force and the displacement. Work is measured in joules (J).

Two consequences worth internalising. A force perpendicular to the motion does no work, since \( \cos 90^{\circ} = 0 \) — which is why the normal contact force never appears in an energy equation. And work is negative when the force opposes the motion, which is how friction enters.

The two energies

\( \text{KE} = \tfrac{1}{2}mv^{2} \)
\( \text{PE} = mgh \)

Both in joules. In the potential energy formula, \(h\) is the change in height, measured vertically — not the distance travelled along a slope. For a body moving distance \(d\) down a slope at angle \( \theta \), the height drop is \( d\sin\theta \).

Kinetic energy depends on the square of the speed. Doubling the speed quadruples the kinetic energy. And because it is squared, kinetic energy has no direction and is never negative — a body moving backwards has positive KE. Forgetting the \( \tfrac12 \), or forgetting to square, are the two most frequent arithmetic slips on this topic.

Conservation of mechanical energy

When the only forces doing work are gravity and the applied forces — that is, with no friction or air resistance — the total mechanical energy is constant:

\[ \text{KE}_{\text{start}} + \text{PE}_{\text{start}} = \text{KE}_{\text{end}} + \text{PE}_{\text{end}} \]

This is often the fastest route to a speed, because it needs no acceleration and no time. For a body falling from rest through height \(h\), \( mgh = \tfrac12 mv^{2} \) gives \( v = \sqrt{2gh} \) — and the mass cancels, so all bodies reach the same speed.

Choose a level for zero potential energy and state it. Any level works; consistency is what matters.

The work–energy principle

With resistance present, mechanical energy is not conserved. The general statement is:

\[ \text{work done by driving force} - \text{work done against resistance} = \text{change in KE} + \text{change in PE} \]

Work done against friction over a distance \(d\) is \( \mu R d \) — and note it depends on the distance travelled, not the displacement, so a body that goes up a slope and comes back down loses friction energy on both legs.

This principle also answers “how much energy is lost in the collision?” questions from M 3: compute the total kinetic energy before and after, and the difference is the loss. Momentum is always conserved in a collision; kinetic energy generally is not.

Power

\( P = \dfrac{\text{work done}}{\text{time}} \)
\( P = Fv \)

Power is measured in watts (W), and 1 kW = 1000 W. The second form is the one used for vehicles: at a given instant, the driving force is \( F = \dfrac{P}{v} \).

That relationship explains why a car has a maximum speed. As \(v\) rises, the driving force available from a fixed power output falls; the car stops accelerating when the driving force has dropped to equal the resistance:

\[ \text{at maximum speed:} \quad \frac{P}{v_{\max}} = \text{resistance} \]

On a hill, the resistance to overcome includes the component of weight along the slope, \( mg\sin\theta \), which is why maximum speed uphill is lower.

✏️Worked example

Take \( g = 10 \) m s\(^{-2}\). (a) A block of mass 4 kg is projected up a rough slope inclined at \( 30^{\circ} \) with initial speed 8 m s\(^{-1}\). The coefficient of friction is 0.2. Use energy methods to find the distance it travels up the slope before coming to rest. (b) A car of mass 900 kg has a maximum power output of 45 kW. Find its maximum speed on a level road against a constant resistance of 750 N.

(a) Let the distance travelled up the slope be \(d\) metres.

The initial kinetic energy is \( \tfrac12(4)(8^{2}) = 128 \) J, and the final kinetic energy is zero.

The height gained is \( d\sin 30^{\circ} = 0.5d \), so the gain in potential energy is \( mgh = 4(10)(0.5d) = 20d \) J.

The normal force is \( R = mg\cos 30^{\circ} = 40(0.8660) = 34.64 \) N, so the work done against friction is \( \mu Rd = 0.2(34.64)d = 6.928d \) J.

All the initial kinetic energy goes into potential energy and friction:

\[ 128 = 20d + 6.928d = 26.928d \;\Longrightarrow\; d = 4.75\ \text{m} \]

(b) At maximum speed the acceleration is zero, so the driving force equals the resistance, 750 N. Using \( P = Fv \):

\[ 45\,000 = 750\,v_{\max} \;\Longrightarrow\; v_{\max} = 60\ \text{m s}^{-1} \]
Check it. In (a), compare with the frictionless case: without friction the block would travel \( \dfrac{128}{20} = 6.4 \) m, so friction reducing that to 4.75 m is the right kind of effect — a noticeable reduction, not a tiny or an overwhelming one. In (b), the mass of the car is not needed at all: at constant maximum speed there is no acceleration, so \( F = ma \) never enters. Unused data in a mechanics question is often deliberate.
Height is not distance along the slope. Using \( mgd \) instead of \( mgd\sin\theta \) for the potential energy is the standard error in (a), and it gives \( d = 2.7 \) m. The energy equation needs the vertical rise, while the friction term needs the distance along the surface — the two are genuinely different lengths in the same equation, which is exactly why this trips people.

📝Practise

Work through these, then reveal the answer. Take \( g = 10 \) m s\(^{-2}\) throughout.

1. A force of 60 N acts at \( 40^{\circ} \) to the horizontal on a box which moves 12 m horizontally. Find the work done by the force.
Work \( = Fd\cos\theta = 60 \times 12 \times \cos 40^{\circ} = 720(0.7660) = 552 \) J. Only the component of the force along the direction of motion does work; the vertical component does none.
2. A ball of mass 0.4 kg is dropped from rest at a height of 5 m. Use energy to find its speed just before impact.
Potential energy lost \( = mgh = 0.4(10)(5) = 20 \) J, all converted to kinetic energy. So \( \tfrac12(0.4)v^{2} = 20 \), giving \( 0.2v^{2} = 20 \), \( v^{2} = 100 \) and \( v = 10 \) m s\(^{-1}\). (Equivalently \( v = \sqrt{2gh} = \sqrt{100} \) — the mass cancels.)
3. A cyclist and bicycle of total mass 80 kg travel at a constant 6 m s\(^{-1}\) up a slope inclined at \( 5^{\circ} \), against a resistance of 20 N. Find the power output.
At constant speed the driving force balances resistance plus the weight component: \( F = 20 + mg\sin 5^{\circ} = 20 + 800(0.08716) = 20 + 69.7 = 89.7 \) N. Then \( P = Fv = 89.7 \times 6 = 538 \) W.
4. A car of mass 1000 kg moving at 20 m s\(^{-1}\) brakes to rest in 50 m on a level road. Find the constant braking force.
Initial KE \( = \tfrac12(1000)(400) = 200\,000 \) J, final KE zero. All of it is removed by the braking force over 50 m: \( F \times 50 = 200\,000 \), so \( F = 4000 \) N. (Equivalently, \( v^{2} = u^{2} + 2as \) gives \( a = -4 \), then \( F = ma = 4000 \) N.)
5. Two particles of mass 3 kg and 2 kg move towards each other at 6 m s\(^{-1}\) and 4 m s\(^{-1}\) and coalesce. Find the kinetic energy lost in the collision.
From M 3, the combined velocity is 2 m s\(^{-1}\). KE before \( = \tfrac12(3)(36) + \tfrac12(2)(16) = 54 + 16 = 70 \) J. KE after \( = \tfrac12(5)(4) = 10 \) J. Energy lost \( = 60 \) J. Note both initial kinetic energies are positive even though the velocities had opposite signs — squaring removes the direction.
6. A train of mass \( 2 \times 10^{5} \) kg has a maximum power of 800 kW. Find its maximum speed on the level against a resistance of 16 000 N, and on a slope of \( \sin\theta = 0.01 \).
Level: \( v_{\max} = \dfrac{P}{F} = \dfrac{800\,000}{16\,000} = 50 \) m s\(^{-1}\). On the slope the driving force must also overcome \( mg\sin\theta = 2\times10^{5}(10)(0.01) = 20\,000 \) N, so the total is \( 36\,000 \) N and \( v_{\max} = \dfrac{800\,000}{36\,000} = 22.2 \) m s\(^{-1}\). The slope more than halves the maximum speed, which is why gradients on railways are kept small.

🔗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.

  • PhET — Energy Skate Park, which shows KE and PE trading off in real time
  • The Physics Classroom — work, energy and power
  • Khan Academy — the work–energy theorem