AS Physics: Work, Energy and Power — Revision Notes

Condensed for the final weeks. For the full explanation, use the Work, Energy and Power study guide, and test yourself with the practice questions.

Equations

work done          W = F s cos(theta)     theta = angle between F and s
kinetic energy     Ek = 1/2 m v^2
grav. potential    Ep = m g h             (near the Earth's surface)
power              P = W / t  =  F v
efficiency         = useful output / total input  x 100%

The cos θ matters: work is done only by the component of force along the displacement. A force perpendicular to motion does zero work — which is why circular motion at constant speed involves no work by the centripetal force.

Conservation of energy

Energy cannot be created or destroyed, only transferred between stores.

Falling body (no resistance):   m g h = 1/2 m v^2   ->   v = sqrt(2 g h)

Mass cancels, so all objects reach the same speed after the same drop in the absence of air resistance.

With resistance: Ep = Ek + work done against resistance, so the impact speed is lower and the difference appears as internal energy — the same reasoning used to find the energy dissipated when a dropped ball rebounds to a lower height than it fell from.

Power in two forms

P = W / t       energy transferred per second
P = F v         useful when force and velocity are both known

P = Fv is the one to reach for in vehicle problems: at constant velocity, the driving force equals total resistance, so P = (resistive force) × v — and the acceleration is zero, since the resultant force is zero.

Deriving Ek

From v² = u² + 2as with u = 0 and F = ma:

v^2 = 2as       ->    s = v^2 / 2a
W = F s = ma x v^2/(2a) = 1/2 m v^2

Being asked to derive rather than quote is common on Paper 2 — know every step, not just the final formula.

Efficiency and dissipation

Most real devices are not 100% efficient, because some energy is dissipated – usually as internal energy to the surroundings – where it becomes too spread out to be useful; that energy is not destroyed, only transferred to a form that can no longer do the intended job. Efficiency can never exceed 100%, but 100% itself isn’t automatically impossible: it depends what the “useful” output is defined as. A heater intended to produce heat can convert essentially all of its electrical input into that useful output, since there’s no other form for the energy to escape to.

Worked example. A pump raises 300 kg of water through 10 m in 25 s, while drawing 1500 W of electrical power.

useful power = mgh / t = 300 x 9.81 x 10 / 25 = 1177 W

efficiency = (useful output / total input) x 100 = (1177 / 1500) x 100 = 78.5%

Worked example. A 0.40 kg ball is dropped from 6.0 m and rebounds to 3.5 m. Find the energy dissipated during the bounce.

GPE lost falling  = mgh1 = 0.40 x 9.81 x 6.0 = 23.5 J
GPE regained on rebound = mgh2 = 0.40 x 9.81 x 3.5 = 13.7 J
energy dissipated = 23.5 - 13.7 = 9.81 J

Exam traps

• Omitting cos θ where force and displacement are not parallel.
• Saying energy is “lost” rather than dissipated or transferred.
• Using the slope length instead of the vertical height in mgh.
• Forgetting that at constant velocity the driving force equals resistance, so acceleration is zero.
• Ek depends on v², so doubling speed quadruples kinetic energy.
• Efficiency above 100% means the useful and total values have been swapped.
• On a rebound problem, forgetting to find GPE at both heights before subtracting — the dissipated energy is a difference, not either value alone.
• Mixing up “useful power output” (found from mgh/t) with “total power input” (given, or drawn from the supply) when substituting into the efficiency formula.

Self-test

1. A 50 N force pulls a crate 8 m at 30° to the horizontal. Find the work done.
2. Derive Ek = ½mv² from the equations of motion.
3. A car travels at constant 25 m/s against 800 N of resistance. Find the engine’s useful power output.
4. Why does a centripetal force do no work?
5. A ball is dropped 20 m. Find its impact speed, ignoring air resistance (g = 9.81).
6. A pump raises 200 kg of water through 8.0 m in 20 s, drawing 1000 W. Find its efficiency.
7. A 0.50 kg ball is dropped from 4.0 m and rebounds to 2.0 m. Find the energy dissipated during the bounce.

Answers: 1. W = 50 × 8 × cos 30° = 346 J. 2. From v² = u² + 2as with u = 0, s = v²/2a; W = Fs = ma × v²/2a = ½mv². 3. At constant velocity the driving force equals resistance, so P = Fv = 800 × 25 = 20 kW. 4. It acts perpendicular to the velocity at every instant, so cos θ = cos 90° = 0 and no work is done. 5. v = √(2 × 9.81 × 20) = 19.8 m/s. 6. Useful power = mgh/t = 200 × 9.81 × 8.0 ÷ 20 = 784.8 W; efficiency = (784.8 ÷ 1000) × 100 = 78.5%. 7. GPE lost = 0.50 × 9.81 × 4.0 = 19.6 J; GPE regained = 0.50 × 9.81 × 2.0 = 9.81 J; energy dissipated = 19.6 − 9.81 = 9.81 J.