A hoist raising a load

8 marks · A-level · original practice case

Write your working and explanations below or on paper. Your responses stay in this page; PhysicsHelp does not submit or save them. Copy anything you want to keep before closing it. Check calculations separately from method marks. Use unrounded working between parts.

The situation

A hoist raises a 50 kg load vertically at constant speed. Use g = 9.8 m s⁻². Electrical input power is 300 W. The model position readings are exact. Neglect changes in kinetic energy.

Height readings
t / s	Height / m
0	0
2	1
4	2
6	3

Part (a)

Determine the upward speed from the readings. [2 marks]

Give the numerical answer to 2 significant figures.

Final value / m s⁻¹

Working

Part (b)

Calculate the useful mechanical output power. [2 marks]

Give the numerical answer to 2 significant figures.

Final value / W

Working

Part (c)

Calculate the efficiency as a percentage. [2 marks]

Give the numerical answer to 2 significant figures.

Final value / %

Working

Part (d)

Explain why the cable tension equals the weight during this motion. [2 marks]

Your explanation

Marking rubric and worked answers

Indicative practice marks. Method marks assess valid working; accuracy marks require the associated valid method. Independent points stand on their own. Tick only points supported by your response. Do not award the same explanation twice. These ticks record your judgement, not an examiner’s result.

Part (a) marking points

Final answer: 0.50 m s⁻¹. Numerical checks assess the final value only; they do not check working or the number of digits typed.

Use the gradient, 3/6. (1 method mark)
Obtain 0.50 m s⁻¹. (1 accuracy mark)

No earlier numerical result is required. Award method credit for valid working and accuracy credit for the stated answer. Accept a defensible equivalent method; wrong physics cannot earn dependent accuracy credit.

Common errors: This reverses distance/time.

Part (b) marking points

Final answer: 250 W. Numerical checks assess the final value only; they do not check working or the number of digits typed.

Use P = mgv. (1 method mark)
Obtain 2.5 × 10² W to 2 significant figures. (1 accuracy mark)

Use the speed from part (a) in P = mgv, with g = 9.8 m s⁻².

Common errors: This is the weight, not power. Multiply by upward speed.

Part (c) marking points

Final answer: 82 %. Numerical checks assess the final value only; they do not check working or the number of digits typed.

Divide useful output power by electrical input power, using unrounded working. (1 method mark)
Obtain 82%. (1 accuracy mark)

Use the useful power from part (b) in efficiency = Puseful/300 × 100%. Follow through a numerical error; reversing useful and input power is a wrong model.

Common errors: This is the efficiency fraction; the question asks for a percentage.

Part (d) marking points

Constant velocity means zero acceleration and zero resultant force. (1 independent mark)
Upward tension balances downward weight. (1 independent mark)

Award each distinct physical point once. Equivalent scientifically correct wording is creditable; a contradictory statement cannot earn the same point.

Report a problem with this case

Related skills

Scope and reference sources

AQA 3.4.1 (selected); OCR A Module 3 (selected). These are shared original exercises; wording, indicative marks and difficulty have not been endorsed by either board.

AQA specification · OCR A specification

Case reference: hoist-power · version 1.1.0 · seed 0. Choose another case