AS Physics: Work, Energy and Power — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Work, Energy and Power revision notes and the full study guide, which covers the Eₖ derivation and further worked examples.

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Questions

1. Define work done, and explain the significance of the cos θ term. [3]

2. State the principle of conservation of energy. [2]

3. Explain why the centripetal force acting on an orbiting satellite does no work. [2]

4. A crate of mass 45 kg is pulled 12 m along a horizontal floor by a rope at 30° to the horizontal with a tension of 210 N.

(a) Calculate the work done by the tension. [3] (b) Explain why the weight does no work in this case. [2]

5. A ball of mass 0.35 kg is dropped from a height of 8.0 m and rebounds to 5.2 m. (g = 9.81 m s⁻²)

(a) Calculate the gravitational potential energy lost during the fall. [2] (b) Calculate the speed just before impact, assuming no air resistance. [3] (c) Calculate the energy dissipated during the bounce. [2] (d) State two forms into which that energy is transferred. [2]

6. A pump raises 250 kg of water through 14 m every minute.

(a) Calculate the useful power output. [3] (b) The pump draws 1200 W. Calculate its efficiency. [2] (c) Suggest one reason the efficiency is below 100%. [1]

7. Starting from work done = force × displacement, derive Eₖ = ½mv² for an object of mass m accelerated from rest to speed v by a resultant force. [3]

8. A 0.50 kg ball is thrown vertically upward at 8.0 m s⁻¹. Using conservation of energy, find the maximum height it reaches, ignoring air resistance. (g = 9.81 m s⁻²) [3]

9. State one condition under which ∆Eₚ = mg∆h is not valid. [1]

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Answers

1. Work done = force × displacement in the direction of the force [1], W = Fs cos θ [1]. The cos θ resolves the force so that only the component along the displacement is counted — a force at 90° to the motion does zero work, however large it is [1].

2. Energy cannot be created or destroyed [1], only transferred from one form to another, so the total energy of a closed system is constant — any apparent loss simply reappears as another form, usually thermal energy [1].

3. The centripetal force acts perpendicular to the velocity at every instant [1], and work requires a component of force along the displacement, so cos 90° = 0 and no energy is transferred to or from the satellite by this force [1].

4. (a) W = Fs cos θ = 210 × 12 × cos 30° [1] [1] = 2182 J ≈ 2.18 kJ [1]. (b) The weight acts vertically while the displacement is horizontal [1], so there is no component of weight along the displacement [1].

5. (a) E = mgh = 0.35 × 9.81 × 8.0 [1] = 27.5 J [1]. (b) ½mv² = mgh, so v = √(2gh) = √(2 × 9.81 × 8.0) [1] [1] = 12.5 m s⁻¹ [1]. (c) GPE at rebound = 0.35 × 9.81 × 5.2 = 17.9 J [1] Energy dissipated = 27.5 − 17.9 = 9.6 J [1]. (d) Thermal energy (heating the ball and floor) [1] and sound [1].

6. (a) E = mgh = 250 × 9.81 × 14 = 34 335 J [1] P = E ÷ t = 34 335 ÷ 60 [1] = 572 W [1]. (b) efficiency = 572 ÷ 1200 [1] = 0.477 or 47.7% [1]. (c) Energy is dissipated by friction in the pump or in the pipes, and as sound and heat [1].

7. From v² = u² + 2as with u = 0: s = v² ÷ 2a [1]. Work done W = Fs = ma × (v² ÷ 2a) [1] = ½mv², so Eₖ = ½mv² [1].

8. At maximum height all kinetic energy has converted to gravitational potential energy: ½mv² = mg∆h [1]. ∆h = v² ÷ 2g = 8.0² ÷ (2 × 9.81) [1] = 3.3 m [1].

9. It applies only in a uniform gravitational field; over large height changes, where g varies appreciably, the general (non-uniform-field) treatment covered in Gravitational Fields must be used instead [1].

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Where marks are usually lost

• Omitting cos θ when the force is at an angle.
• Forgetting to convert per-minute figures to per-second for power.
• Saying weight does work when motion is horizontal.
• Not stating that dissipated energy is transferred, not destroyed.
• Skipping a step in the Eₖ = ½mv² derivation — examiners expect every line, from v² = u² + 2as through to the final substitution, not just the final formula.
• In a maximum-height problem, forgetting that all the kinetic energy converts to potential energy only at the very top, where the vertical velocity is momentarily zero.
• Treating P = Fv as the general definition of power — the general definition is always P = W/t. P = Fv is a derived instantaneous-power result that holds generally (including while accelerating), but it is easiest to use in a constant-velocity problem against a resistive force, where the driving force is simply equal to the resistive force.
• Applying ∆Eₚ = mg∆h to a large change in height without checking that g is approximately constant over that range.