AS & A Level · Practice questions · Kinetic and potential energy, the work-energy principle

Keep the account.

Six original Cambridge-style questions on kinetic and potential energy, conservation of energy and the work-energy principle, with full worked solutions.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Start + work in = end + work out.

KE \(= \tfrac{1}{2}mv^2\) at the start and the end.
PE change \(= mg \times\) vertical height change (not distance along a slope).
Work in: driving forces. Work out: friction and resistance (\(\mu R \times d\)).
Equation: initial KE \(+\) PE \(+\) work in \(=\) final KE \(+\) PE \(+\) work out.
Curved paths: energy is the only method you need.

Recall

[2 marks]

A ball of mass 0.5 kg moves at \(8\ \text{m s}^{-1}\) at a height of 3 m above the ground. Find its kinetic energy and its potential energy relative to the ground.

KE \(= \tfrac{1}{2}(0.5)(64) = 16\) J. ✓
PE \(= 0.5 \times 10 \times 3 = 15\) J. ✓

Calculation

[2 marks]

A child of mass 30 kg slides from rest down a smooth curved slide, descending a vertical height of 4 m. Find the speed at the bottom.

\(30(10)(4) = \tfrac{1}{2}(30)v^2\), so \(v^2 = 80\). ✓
\(v = 8.94\ \text{m s}^{-1}\). ✓

Calculation

[3 marks]

A stone is thrown from the top of a cliff 20 m above the sea with speed \(15\ \text{m s}^{-1}\). Ignoring air resistance, find the speed with which it hits the sea.

\(\tfrac{1}{2}v^2 = \tfrac{1}{2}(15^2) + 10(20)\). ✓
\(v^2 = 225 + 400 = 625\). ✓
\(v = 25\ \text{m s}^{-1}\). ✓

Analysis

[6 marks]

A particle of mass 2 kg is projected at \(10\ \text{m s}^{-1}\) up a line of greatest slope of a rough plane with \(\sin\alpha = 0.6\). The coefficient of friction is 0.25. Use energy to find (a) the distance it travels up the plane, (b) its speed when it slides back to its starting point.

Friction \(= 0.25 \times 20 \times 0.8 = 4\) N. ✓
(a) Up a distance \(d\): \(\tfrac{1}{2}(2)(100) = 20(0.6)d + 4d\), so \(100 = 16d\) and \(d = 6.25\) m. ✓
(b) Down: PE lost \(= 12 \times 6.25 = 75\) J; work against friction \(= 4 \times 6.25 = 25\) J. ✓
\(\tfrac{1}{2}(2)v^2 = 75 - 25 = 50\), so \(v = 7.07\ \text{m s}^{-1}\). ✓

Analysis

[4 marks]

A car of mass 1000 kg freewheels along a level road, its speed falling from \(25\ \text{m s}^{-1}\) to \(15\ \text{m s}^{-1}\) over 200 m. Assuming a constant resistance, find the resistance and how much further the car travels before stopping.

KE lost \(= \tfrac{1}{2}(1000)(625 - 225) = 200\,000\) J, so \(R = 1000\) N. ✓
Remaining KE \(= \tfrac{1}{2}(1000)(225) = 112\,500\) J. ✓
Further distance \(= 112\,500 \div 1000 = 112.5\) m. ✓

Analysis

[6 marks]

A roller-coaster car of mass 500 kg starts from rest at A, 30 m above the ground. It passes B, 12 m above the ground, then C, 20 m above the ground. Between A and B the resistances do 15 000 J of work against the car; between B and C the track may be taken as smooth with no resistance. Find the speed of the car at B and at C.

A to B: PE lost \(= 500 \times 10 \times 18 = 90\,000\) J. ✓
\(\tfrac{1}{2}(500)v_B^2 = 90\,000 - 15\,000 = 75\,000\), so \(v_B^2 = 300\) and \(v_B = 17.3\ \text{m s}^{-1}\). ✓
B to C: rises 8 m, so \(\tfrac{1}{2}v_C^2 = \tfrac{1}{2}(300) - 10(8)\). ✓
\(v_C^2 = 140\), so \(v_C = 11.8\ \text{m s}^{-1}\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.