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Questions on work, energy and power
Work is the energy transferred when a force moves its point of application, and only the force component along the motion does work. Where the force varies, the work is the area under the force-displacement graph. Power is the rate of that transfer, P = Fv for a force along the motion, and efficiency is the useful fraction of the input.
20 original questions · 59 marks · the work, energy and power notes · Mechanics
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A force of 20 N moves an object 3.0 m in the direction of the force. Calculate the work done.
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W = Fs = 20 × 3.0 (1)
W = 60 J (1)Define power and state its SI unit.
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Power is the rate of doing work, or of transferring energy: P = work done / time taken (1). Unit: the watt, W, equal to one joule per second (1).State the equations for kinetic energy and for the change in gravitational potential energy near the Earth's surface.
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Kinetic energy = ½mv2 (1); change in gravitational potential energy = mgΔh (1).A shopper carries a bag at constant height across a level floor. State and explain the work done on the bag by the upward force from the shopper's hand.
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The force is vertical and the displacement horizontal, so the angle between them is 90° (1). W = Fs cos 90° = 0: the force does no work on the bag. (1)A kettle transfers energy at a rate of 2.5 kW. Calculate the energy it transfers in 3.0 minutes.
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P = 2500 W and t = 180 s (1). E = Pt = 2500 × 180 = 4.5 × 105 J. (1)State the quantity represented by the area under a force-displacement graph.
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The work done by the force. (1)A force of 50 N pulls a crate 4.0 m along the ground, acting at 30° above the horizontal. Calculate the work done by the force.
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W = Fs cosθ (1)
W = 50 × 4.0 × cos 30° (1)
W = 173 J (1)A 2.0 kg object moves at 6.0 m s−1. Calculate (a) its kinetic energy and (b) the distance it travels before stopping if a constant 12 N resistive force acts on it.
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(a) KE = ½ × 2.0 × 6.02 (1)
KE = 36 J (1)
(b) Work done against the force removes this energy: s = KE/F = 36/12 (1)
s = 3.0 m (1)A motor raises a 20 kg load through a height of 5.0 m in 4.0 s at steady speed. Taking g = 9.81 m s−2, calculate the useful output power of the motor.
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P = mgΔh/t (1)
P = (20 × 9.81 × 5.0)/4.0 (1)
P = 245 W (1)A machine has an input power of 800 W and a useful output power of 600 W. Calculate its efficiency.
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Efficiency = useful output / input = 600/800 (1)
Efficiency = 0.75, or 75% (1)The force needed to stretch a spring increases linearly from zero to 12 N as its extension increases from zero to 80 mm. Determine the work done in stretching the spring.
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Work done = area under the force-extension graph (1) = ½ × 12 × 0.080 (1) = 0.48 J. (1)A sprinter of mass 62 kg accelerates from rest to 9.0 m s−1 in 3.5 s. Calculate the average useful power she develops.
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Energy transferred = kinetic energy gained = ½ × 62 × 9.02 = 2511 J (1). P = E/t = 2511/3.5 (1) = 717 W (about 720 W). (1)A van of mass 900 kg climbs a straight road at a steady 15 m s−1. The road rises 1.0 m for every 20 m travelled along it. Taking g = 9.81 m s−2, calculate the additional output power the engine must supply against gravity, compared with driving at the same speed on a level road.
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Height gained per second = 15/20 = 0.75 m (1). Extra power = mg × height gained per second = 900 × 9.81 × 0.75 (1) = 6.6 kW. (1)A car engine develops a useful output power of 40 kW while the car travels at a steady 25 m s−1. Calculate the driving force provided by the engine.
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P = Fv (1)
F = P/v = 40000/25 (1)
F = 1600 N (1)A 10 kg box is pushed at steady speed a distance of 4.0 m up a slope inclined at 30° to the horizontal, against a constant frictional force of 15 N. Taking g = 9.81 m s−2, calculate (a) the gain in gravitational potential energy, (b) the work done against friction, and (c) the force applied parallel to the slope.
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Height gained = 4.0 sin 30° = 2.0 m (1)
(a) ΔGPE = mgΔh = 10 × 9.81 × 2.0 (1)
ΔGPE = 196 J (1)
(b) Wfriction = 15 × 4.0 = 60 J (1)
(c) Total work = 256 J over 4.0 m, so F = 256/4.0 = 64 N (1)Explain why the efficiency of a real machine is always less than 100%, and where the 'lost' energy goes.
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Some input energy is always transferred to unwanted forms (1), mainly heat from friction and air resistance, and sound (1). This energy is dissipated to the surroundings, so the useful output is always less than the input (1).A borehole pump raises 120 kg of water through a vertical height of 15 m every minute; the water leaves the outlet with negligible kinetic energy. The pump's electrical input power is 500 W, and the manufacturer claims an efficiency of at least 65%. Taking g = 9.81 m s−2, deduce whether the pump meets the manufacturer's claim.
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Useful output power = mgΔh/t = 120 × 9.81 × 15/60 (1) = 294 W (1). Efficiency = 294/500 = 0.59 (59%) (1). This is below 65%, so the pump does not meet the claim. (1)A resultant force acts on a 0.50 kg trolley, initially at rest, along a straight track. The force is a constant 6.0 N for the first 2.0 m of travel, then decreases linearly to zero over the next 2.0 m. Determine the speed of the trolley after it has travelled 4.0 m.
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Work done = area under the force-displacement graph (1): 6.0 × 2.0 + ½ × 6.0 × 2.0 = 12 + 6.0 = 18 J (1). ½mv2 = 18 J (1), so v = √(2 × 18/0.50) = 8.5 m s−1. (1)Estimate the useful power a student develops against gravity when running up a flight of stairs. State the estimated values you use.
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Sensible estimates, e.g. mass 60 kg and a height gain of 3.0 m in about 3 s (1). P = mgΔh/t = 60 × 9.81 × 3.0/3.0 (1) ≈ 590 W, i.e. a few hundred watts. (1)A sledge of mass 8.0 kg starts from rest and is pulled along a straight horizontal track. The graph shows the resultant force on the sledge against its displacement along the track. Determine the total work done on the sledge over the 10 m shown, its speed after 10 m, and the displacement at which it is moving fastest.
The resultant force on the sledge against displacement. Part of the line lies below the displacement axis. Mark scheme
The work done is the area between the line and the displacement axis, with area below the axis counted as negative because the force there opposes the motion (1)
Above the axis the area is a trapezium of height 40 N whose parallel sides are the base, from 0 to the 8.0 m at which the line crosses the axis, and the flat top from 2.0 m to 5.0 m: ½ × (8.0 + 3.0) × 40 = 220 J (1)
Below the axis the area is 20 × 2.0 = 40 J, so the total work done is 220 − 40 = 180 J (1)
The work done by the resultant force equals the gain in kinetic energy, so ½ × 8.0 × v2 = 180 and v = 6.7 m s−1 (1)
The sledge is moving fastest at a displacement of 8.0 m, where the resultant force changes sign and begins to remove kinetic energy (1)
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