OxfordAQA International AS and A-level Mathematics: Work, energy and power (9660) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section M2.6 (Work and energy) of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2. The specification states that this content is part of International A-level (A2) Unit M2 (Mechanics) and is not examined at International AS. Take g = 9.8 m s⁻² and give answers to 3 significant figures unless a question says otherwise.

Links: study guide, revision notes, course hub, checklist and free diagnostics.

Questions

1. A person pulls a lawn mower 18 m along level ground by pushing on its handle with a constant force of 60 N. The force acts at 30° to the horizontal. Find the work done by the force. [2]

2. A ball of mass 0.25 kg is thrown vertically upwards with speed 14 m s⁻¹. Find its initial kinetic energy and, using conservation of energy, the greatest height it reaches above the starting point. [3]

3. A car of mass 1200 kg slows from 25 m s⁻¹ to 10 m s⁻¹ while travelling 70 m along a straight horizontal road. Assuming a constant braking force and no other resistance, find the size of the braking force. [4]

4. A 2 kg particle starts from rest on a smooth slope and slides until it is 3.6 m vertically lower than where it began. Find its speed at that point. [3]

5. A 2 kg block is launched at 7.5 m s⁻¹ straight up a rough ramp, which makes an angle α with the horizontal, where sin α = 0.6. While the block moves, the friction on it is a constant 3.24 N.

(a) Find how far up the plane the block travels before it comes to rest. [4] (b) The block then slides back down, with the friction unchanged at 3.24 N. Find its speed when it returns to its starting point. [3]

6. A car travels at a constant speed of 25 m s⁻¹ along a straight horizontal road with the engine working at 30 kW. The resistance to motion is constant.

(a) Find the resistance. [2] (b) The car has mass 1600 kg. Find its acceleration when its speed is 15 m s⁻¹, assuming the power and the resistance are unchanged. [3]

7. A cyclist and bicycle have total mass 80 kg. The cyclist rides at a steady 4 m s⁻¹ up a straight road inclined at an angle θ to the horizontal, where sin θ = 1/14. The resistance to motion is a constant 25 N.

(a) Find the power the cyclist develops. [3] (b) The cyclist’s greatest sustained power is 400 W. Find the greatest steady speed that can be kept up on this road. [2]

8. A crane lifts a load of mass 450 kg from rest on the ground. After the load has risen 6 m it is moving upwards at 2 m s⁻¹. Assuming there is no resistance, find the work done by the crane cable on the load. [4]

9. A pendulum bob of mass 0.4 kg hangs from a light inextensible string of length 1.2 m. It is released from rest with the string taut and at 60° to the downward vertical. Find the speed of the bob at its lowest point. [3]

10. A car of mass 1000 kg has a maximum engine power of 40 kW. The resistance to motion is a constant 800 N on every road.

(a) Find the maximum speed of the car on a level road. [2] (b) The car climbs a straight road inclined at an angle β to the horizontal, where sin β = 0.05. Find its greatest steady speed. [3] (c) On the level road the car works at maximum power. Find its acceleration when its speed is 15 m s⁻¹. [3]

11. A block of mass 2.5 kg is initially at rest on a rough horizontal table. The coefficient of friction between block and table is 0.2. A constant force of 20 N, at 35° above the horizontal, pulls the block 8 m along the table.

(a) Find the speed of the block after the 8 m. [5] (b) The pulling force is now removed and the block slides on until it stops. Find the further distance it travels. [3]

12. A ball is kicked from ground level with speed 20 m s⁻¹ at 40° above the horizontal. Ignore air resistance.

(a) Find the speed of the ball when it is 6 m above the ground. [2] (b) Find the greatest height reached by the ball. [3]

Answers

1. Work = Fd cos θ = 60 × 18 × cos 30° [1] = 935.3… J, so 935 J [1]. [2] Examiner insight: write the formula with your numbers substituted before the final value; a visible correct setup shows your method even if the arithmetic slips.

2. KE = ½ × 0.25 × 14² [1] = 24.5 J. At the top all this is GPE: 24.5 = 0.25 × 9.8 × h [1], so h = 10 m [1]. [3] Examiner insight: the mass cancels in the energy equation, so check whether you can use v² = 2gh as a quick cross-check: 196 ÷ 19.6 = 10.

3. Work done by braking force = change in KE: −F × 70 [1] = ½ × 1200 × (10² − 25²) [1] = −315 000 [1]. So F = 4500 N [1]. [4] Examiner insight: a negative sign on the work (force opposite to motion) must match a negative change in KE; a positive F in a negative equation is a sign that you have lost a minus.

4. The normal reaction does no work, so mechanical energy is conserved: ½mv² = mgh, so v² = 2gh [1] = 2 × 9.8 × 3.6 = 70.56 [1]. v = 8.4 m s⁻¹ [1]. [3] Examiner insight: say why energy is conserved (smooth surface, normal reaction perpendicular to motion).

5. (a) Initial KE = ½ × 2 × 7.5² = 56.25 J [1]. Resisting forces along the slope: weight component 2 × 9.8 × 0.6 = 11.76 N, plus friction 3.24 N, so 15.00 N in total [1]. Work-energy: 56.25 = 15 d [1]. d = 3.75 m [1]. (b) Moving down, gravity helps and friction opposes: net force 11.76 − 3.24 = 8.52 N, work over 3.75 m = 31.95 J [1]. ½ × 2 × v² = 31.95 [1]. v² = 31.95, so v = 5.65 m s⁻¹ [1]. [7] Examiner insight: the weight component acts down the slope in both directions of travel, but friction reverses. Writing the direction of each force beside its value helps stop the sign error in part (b).

6. (a) At constant speed the driving force equals the resistance [1]. R = P/v = 30 000/25 = 1200 N [1]. (b) Driving force F = 30 000/15 = 2000 N [1]. Newton’s second law: 2000 − 1200 = 1600a [1], so a = 0.5 m s⁻² [1]. [5] Examiner insight: convert 30 kW to 30 000 W before dividing. In part (b) the driving force at the new speed is not 1200 N, so reusing part (a)’s value is a common route to a wrong answer.

7. (a) Constant speed so driving force F = 25 + 80 × 9.8 × (1/14) [1] = 25 + 56 = 81 N [1]. P = Fv = 81 × 4 = 324 W [1]. (b) The driving force needed at constant speed is still 81 N [1]. v = 400/81 = 4.94 m s⁻¹ [1]. [5] Examiner insight: the force needed at a steady speed on this hill does not depend on the speed, so part (b) follows quickly from part (a).

8. The cable does work equal to the gain in GPE plus the gain in KE (no resistance) [1]. GPE gain = 450 × 9.8 × 6 = 26 460 J [1]. KE gain = ½ × 450 × 2² = 900 J [1]. Work = 26 460 + 900 = 27 360 J, which is 27 400 J to 3 s.f. [1]. [4] Examiner insight: the cable tension is not constant here, so a method that uses tension × distance without a stated assumption is risky. The energy route avoids needing the tension.

9. Vertical drop = 1.2 − 1.2 cos 60° = 0.6 m [1]. The string tension is perpendicular to the motion, so it does no work and energy is conserved: ½v² = 9.8 × 0.6 [1]. v² = 11.76, so v = 3.43 m s⁻¹ [1]. [3] Examiner insight: the drop is measured from the starting height to the lowest point, not along the arc.

10. (a) At maximum speed the driving force equals the resistance, 800 N [1]. v = 40 000/800 = 50 m s⁻¹ [1]. (b) Driving force F = 800 + 1000 × 9.8 × 0.05 [1] = 1290 N [1]. v = 40 000/1290 = 31.0 m s⁻¹ [1]. (c) F = 40 000/15 = 2666.7… N [1]. 2666.7 − 800 = 1000a [1]. a = 1.87 m s⁻² [1]. [8] Examiner insight: in part (b) the weight component must be included in the force balance.

11. (a) Vertically: R + 20 sin 35° = 2.5 × 9.8 [1]. R = 24.5 − 11.47… = 13.03… N [1]. The block is moving, so friction = 0.2R = 2.606… N [1]. Work-energy: 20 × 8 × cos 35° − 2.606… × 8 = ½ × 2.5 × v² [1], i.e. 131.06 − 20.85 = 110.22 = 1.25v². So v = 9.39 m s⁻¹ [1]. (b) Friction is now 0.2 × 2.5 × 9.8 = 4.9 N [1]. Work-energy: −4.9 × d = −110.22 [1], so d = 22.5 m [1]. [8] Examiner insight: the upward component of the pull reduces R, so using R = mg in part (a) is a classic error. In part (b) the normal reaction is mg once the pull has gone.

12. (a) Energy is conserved: ½v² = ½ × 20² − 9.8 × 6 [1]. v² = 400 − 117.6 = 282.4, so v = 16.8 m s⁻¹ [1]. (b) At the greatest height the velocity is horizontal, with speed 20 cos 40° = 15.32… m s⁻¹ [1]. Energy: ½(400) − ½(15.32…²) = 9.8H [1]. 200 − 117.36 = 82.64, so H = 8.43 m [1]. [5] Examiner insight: a projectile still has its horizontal speed at the highest point of its path. Using KE = 0 at the highest point gives a height of about 20.4 m, which is far too large.

Where marks are usually lost

• Leaving out cos θ in work done, or using the angle to the vertical.
• Using the slope length rather than the vertical height in mgh.
• Missing the weight component along the slope in a work-energy equation.
• Sign errors when friction does negative work.
• Using conservation of energy when friction or an engine is present.
• Forgetting to convert kW to W before using P = Fv.
• Putting the net force into P = Fv rather than the force produced by the engine.
• Treating a projectile as stationary at its highest point.
• Using R = mg when the force has a vertical component.
• Rounding in the middle of a calculation, which shifts the final digit.

Next steps

• Reread the study guide and the revision notes.
• Return to the OxfordAQA A-level Mathematics hub and tick off outcomes on the checklist.
• Try all free 10-minute diagnostics at /diagnostics/.
• Book a free trial class at /trial/.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, section M2.6 (Work and energy). Published by OxfordAQA for International AS exams from May/June 2018 and International A-level exams from May/June 2019.