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Work, energy and power questions
Energy methods restate Newton's laws in terms of work rather than force. The work done by the forces acting equals the change in kinetic energy, and where no resistance acts the total mechanical energy stays the same.
7 original questions · 30 marks · the work, energy and power notes · Further Mechanics 1
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State the formula for work done by a constant force, and explain why the normal reaction never appears in an energy equation.
Worked answer
Work = Fd cos θ, where θ is the angle between the force and the displacement. The normal reaction acts at right angles to the surface and so at right angles to the motion along it, making cos θ = 0 and the work done zero. B1 for the formula, B1 for the normal reaction doing no work.A force of 40 N acts at 25° to the direction of motion while an object moves 12 m. Find the work done.
Worked answer
Work = 40 × 12 × cos25° = 480 × 0.9063 = 435 J to three significant figures. M1 for 40 × 12 × cos25°, A1 for 435 J. Ignoring the angle would give 480 J, an overestimate of about 10%.A particle of mass 60 kg slides from rest 10 m down a slope at 20° to the horizontal, with coefficient of friction 0.3. Find its speed at the bottom, taking g = 9.8 m/s².
Worked answer
Work by gravity = 60(9.8)(10)sin20° = 2011 J. Normal reaction = 60(9.8)cos20° = 553 N, so friction is 166 N and the work against it is 1658 J. By the work-energy principle the kinetic energy gained is 2011 − 1658 = 353.5 J, so ½(60)v² = 353.5, v² = 11.78 and v = 3.43 m/s. M1 for the work done by gravity, A1 for 2011 J, M1 for the normal reaction and friction, A1 for 1658 J, M1 for the work-energy principle, A1 for 3.43 m/s. Resolve perpendicular to the slope for the normal reaction before touching the friction, since R = mg cos20° rather than mg.A car of mass 900 kg works at 12 kW. At 20 m/s the resistance is 400 N. Find the acceleration then, and the maximum speed if the resistance stays constant.
Worked answer
Driving force = 12000/20 = 600 N. Resultant = 600 − 400 = 200 N, so a = 200/900 = 0.222 m/s². At maximum speed the acceleration is zero, so 12000/v = 400 and v = 30 m/s. M1 for P = Fv, A1 for 600 N, M1 for the resultant, A1 for the acceleration, B1 for 30 m/s.Explain why a car's acceleration falls as it speeds up, even though the engine works at constant power.
Worked answer
With P fixed, the driving force P/v falls as v rises. The resistance meanwhile grows, so the resultant force shrinks from both ends and the acceleration falls with it. At the speed where the driving force has dropped to equal the resistance, the acceleration reaches zero and the car can go no faster. B1 for the driving force falling, B1 for the resistance rising, B1 for the resultant shrinking to zero.A cyclist and bicycle of total mass 80 kg ride at a steady 6 m/s up a slope of 1 in 20 against a resistance of 25 N. Find the power developed.
Worked answer
A slope of 1 in 20 means the component of weight along the slope is 80(9.8)/20 = 39.2 N. At steady speed the driving force balances that plus the resistance: 39.2 + 25 = 64.2 N. Power = Fv = 64.2 × 6 = 385 W to three significant figures. M1 for the component of weight, A1 for 39.2 N, M1 for P = Fv, A1 for 385 W.A car of mass 1000 kg has an engine working at a constant 20 kW. The resistance to motion is 500 N at all speeds. Find the greatest speed on a level road, the greatest speed up a hill of 1 in 14, and the acceleration up that hill at the moment when the speed is 10 m/s. Take g = 9.8 m/s².
Worked answer
On the level the greatest speed comes when the driving force has fallen to the resistance, so 20000/v = 500 and v = 40 m/s.
A hill of 1 in 14 has sin θ = 1/14, so the component of weight down the slope is 1000(9.8)/14 = 700 N. Climbing at a steady speed needs a driving force of 500 + 700 = 1200 N, so 20000/v = 1200 and v = 16.7 m/s.
At 10 m/s the driving force is 20000/10 = 2000 N. Newton's second law along the slope gives 2000 − 500 − 700 = 1000a, so a = 0.8 m/s².
M1 for 20000/v = 500, A1 for 40 m/s, M1 for the component of weight, A1 for 700 N, M1 for the driving force needed, A1 for 16.7 m/s, M1 for Newton's second law, A1 for the acceleration.
Power is fixed, not force, so the driving force must be recomputed at every speed. Include the component of weight on the hill and omit it on the level, and never carry the maximum-speed driving force into an acceleration calculation.
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