AS Forces and MotionEnergy & Power 3.3.1(a)3.3.2(a)3.2.1(a)3.2.2(a)3.4.1.2(a)

~15 min Difficulty: 5/10 10 marks

Prior knowledge Newton’s second lawwork donepower

Problem structure

Part (a): acceleration from tension and weight (3 marks). Part (b): power at constant speed (2 marks). Part (c): work done during deceleration (3 marks). Part (d): power comparison (2 marks).

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A lift of total mass 6.0 × 10² kg is accelerated vertically upwards from rest by a cable tension of 6.6 × 10³ N. (a) Calculate the acceleration of the lift. (3) (b) The lift reaches a constant speed of 3.0 m s⁻¹. Calculate the power output of the motor when the lift is travelling at this constant speed. (2) (c) The lift then decelerates uniformly to rest over a vertical distance of 4.5 m. Calculate the work done by the cable tension during this deceleration. (3) (d) State and explain whether the power output of the motor is greater when the lift is accelerating upwards or when it is moving upwards at constant speed. (2)

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(a) Resultant upward force = T − mg = 6.6 × 10³ − (600 × 9.81) = 6.6 × 10³ − 5.89 × 10³ = 714 N. Using F = ma: a = 714 / 600 = 1.19 m s⁻² ≈ 1.2 m s⁻² upwards. (b) At constant speed, tension equals weight = 5.89 × 10³ N. Power = tension × speed = 5.89 × 10³ × 3.0 = 1.77 × 10⁴ W ≈ 1.8 × 10⁴ W. (c) For deceleration from 3.0 m s⁻¹ to rest over 4.5 m: v² = u² + 2as ⇒ 0 = 3.0² + 2a × 4.5 a = −9.0 / 9.0 = −1.0 m s⁻² (magnitude 1.0 m s⁻²). Resultant downward force = ma = 600 × 1.0 = 600 N. Tension during deceleration = mg − ma = 5.89 × 10³ − 600 = 5.29 × 10³ N. Work done by tension = T × d = 5.29 × 10³ × 4.5 = 2.38 × 10⁴ J ≈ 2.4 × 10⁴ J. (Positive because tension acts in the direction of motion.) (d) Power is greater during upward acceleration. The motor must provide tension greater than the weight to produce an upward resultant force, whereas at constant speed the tension only equals the weight.

Marking guidance

(a) [3] - Correct resultant force expression (1) - Correct value of resultant force (1) - Correct acceleration 1.2 m s⁻² (1) (b) [2] - Recognises tension equals weight at constant speed (1) - Correct power 1.8 × 10⁴ W (1) (c) [3] - Correct deceleration magnitude (1) - Correct tension during deceleration (1) - Correct work done 2.4 × 10⁴ J (1) (d) [2] - States greater during acceleration (1) - Explanation comparing tension to weight (1)