Flashcards · 9702 · AS Topic 5

Work, energy and power (AS Level) flashcards

12 cards on AS Topic 5, subtopics 5.1 to 5.2. Pick a subtopic or study them all.

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Work done: Force × displacement in the direction of the force: W = Fs cos θ, where θ is the angle between F and s. Unit J.
Why does a force perpendicular to the motion do no work?: cos 90° = 0: no component of the force acts along the displacement. So a centripetal force does no work.
Principle of conservation of energy: Energy cannot be created or destroyed, only transferred from one store to another. The total energy of a closed system is constant.
Efficiency: Efficiency = useful energy (or power) output ÷ total energy (or power) input, × 100%. It can never exceed 100%.
Power, and derive P = Fv: Power is work done per unit time, P = W ÷ t (unit W). For a constant force: P = Fs ÷ t = Fv.
A car moves at a constant 25 m s⁻¹ against 800 N of resistance. Find the engine’s useful power.: At constant velocity the driving force equals the resistance: P = Fv = 800 × 25 = 20 kW.

Derive ΔEp = mgΔh: Lifting a mass at steady speed needs a force equal to its weight, mg. Work done = force × distance = mg × Δh, which is stored as gravitational potential energy (in a uniform field).
Derive Ek = ½mv² from the equations of motion: Accelerate from rest with constant force F: v² = 2as, so s = v² ÷ 2a. W = Fs = ma × v² ÷ 2a = ½mv².
A ball is dropped 20 m. Find its impact speed (no air resistance).: mgh = ½mv², so v = √(2gh) = √(2 × 9.81 × 20) = 19.8 m s⁻¹. The mass cancels.
The speed of a car doubles. What happens to its kinetic energy?: It quadruples, because Ek ∝ v².
An object slides down a ramp. Which height goes in mgΔh?: The vertical height fallen, not the length of the slope.
A 0.50 kg ball is dropped from 4.0 m and rebounds to 2.0 m. Find the energy dissipated in the bounce.: GPE lost = 0.50 × 9.81 × 4.0 = 19.6 J. GPE regained = 0.50 × 9.81 × 2.0 = 9.81 J. Dissipated = 19.6 − 9.81 = 9.81 J.