Kinetic and potential energy turn into each other as a particle moves up and down. When only gravity does work the total stays fixed; when friction or a driving force acts, the work they do accounts exactly for the change.
\(\text{KE} = \tfrac{1}{2}mv^2\) and gravitational \(\text{PE} = mgh\). If only gravity does work, \(\text{KE} + \text{PE}\) is constant. Otherwise use the work-energy principle: initial energy \(+\) work done by driving forces \(=\) final energy \(+\) work done against resistances.
Fig. 1: a child slides from rest down a smooth curved slide that drops 4 m. Only gravity does work, so mg(4) = ½mv² whatever the shape of the slide
Section 01
Two kinds of energy.
In the lab, release the ball on the slope with friction set to zero and watch the bar chart: potential energy drains into kinetic energy and the total bar stays the same height. Then add friction and see a third bar, the work done against friction, take its share.
Kinetic energy \(= \tfrac{1}{2}mv^2\): depends on speed only, never negative, never on direction.
Gravitational potential energy \(= mgh\): measured from any level you choose; only CHANGES in PE matter.
Conservation: with no work done by forces other than gravity (smooth surfaces, no resistance), \(\text{KE} + \text{PE}\) is constant.
The path does not matter: on a smooth curved slide only the change in height counts. That is why energy beats suvat for curved paths.
Worked example A smooth curved slide (Fig. 1)
A child of mass 30 kg starts from rest at the top of a smooth slide and descends a vertical height of 4 m. Find the speed at the bottom.
PE lost \(= 30 \times 10 \times 4 = 1200\) J.
KE gained: \(\tfrac{1}{2}(30)v^2 = 1200\), so \(v^2 = 80\).
\(v = 8.94\ \text{m s}^{-1}\).
Answer: \(8.94\ \text{m s}^{-1}\). The mass cancels: \(v = \sqrt{2gh}\).
Worked example Thrown from a height
A ball is thrown with speed \(12\ \text{m s}^{-1}\) from a point 5 m above the ground. Find its speed when it hits the ground, whatever the direction of throw (ignore air resistance).
Per unit mass: \(\tfrac{1}{2}v^2 = \tfrac{1}{2}(12^2) + 10 \times 5\).
\(v^2 = 144 + 100 = 244\).
\(v = 15.6\ \text{m s}^{-1}\).
Answer: \(15.6\ \text{m s}^{-1}\), the same for any direction of throw.
Your turn
A particle slides from rest down a smooth curved track and reaches the bottom at \(10\ \text{m s}^{-1}\). Find the vertical height it descended.
\(gh = \tfrac{1}{2}v^2\), so \(10h = 50\) and \(h = 5\) m.
Examiner trap
Adding speeds instead of energies. The ball thrown at 12 and gaining 10 from the fall does NOT land at 22: energies add, so \(v^2 = 144 + 100\).
Section 02
The work-energy principle.
When friction, resistance or a driving force does work, keep an energy account. Energy at the start plus work done by driving forces equals energy at the end plus work done against resistances.
Worked example Down a rough slope
A particle of mass 2 kg slides 5 m down a line of greatest slope of a rough plane with \(\sin\alpha = 0.6\), \(\mu = 0.25\). Its speed at the top is \(3\ \text{m s}^{-1}\). Find its speed after the 5 m.
PE lost \(= 20 \times 5 \times 0.6 = 60\) J.
\(R = 20 \times 0.8 = 16\) N, friction \(= 4\) N, work against friction \(= 4 \times 5 = 20\) J.
A car of mass 1000 kg freewheels on a level road, slowing from \(25\ \text{m s}^{-1}\) to \(15\ \text{m s}^{-1}\) over 200 m. Find the constant resistance.
KE lost \(= \tfrac{1}{2}(1000)(25^2 - 15^2) = 500 \times 400 = 200\,000\) J.
All of it is work done against the resistance: \(R \times 200 = 200\,000\).
Answer: \(R = 1000\) N
Examiner trap
\(\tfrac{1}{2}m(25 - 15)^2\) is NOT the change in KE. Square each speed first: \(\tfrac{1}{2}m(25^2 - 15^2)\).
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
Find the kinetic energy of a 4 kg particle moving at \(5\ \text{m s}^{-1}\).
Quick check+10 XP
A 3 kg mass is raised 2 m. Find the gain in gravitational potential energy.
Quick check+10 XP
A particle slides from rest down a smooth curve through a vertical height of 5 m. Find its speed at the bottom.
Quick check+10 XP
When is the total of kinetic and potential energy constant?
Section 03
Driving uphill.
With a driving force, its work goes in on the left of the account. Speed may rise or fall: if the particle slows down, the KE term on the right is smaller than on the left.
Worked example A cyclist on a hill
A cyclist and bicycle of total mass 80 kg ride 300 m along a road that rises 15 m vertically. The speed falls from \(8\ \text{m s}^{-1}\) to \(5\ \text{m s}^{-1}\), and there is a constant resistance of 20 N. Find the work done by the cyclist.
KE at start \(= \tfrac{1}{2}(80)(64) = 2560\) J; at end \(= \tfrac{1}{2}(80)(25) = 1000\) J.
PE gained \(= 80 \times 10 \times 15 = 12\,000\) J.
Work against resistance \(= 20 \times 300 = 6000\) J.
\(2560 + W = 1000 + 12\,000 + 6000\), so \(W = 16\,440\) J.
Answer: 16.4 kJ (3 s.f.)
Examiner trap
Watch the sign of the KE change. Here the cyclist SLOWS down, so the lost KE (1560 J) helps: \(W = 18\,000 - 1560\), not \(18\,000 + 1560\).
Stage 2 · Exam
Exam-style questions
Unlocks once the four checks above are done. Worth more XP, written to AS & A Level Paper 1 to 6 standard.
Finish the four checks above to unlock the exam questions
Exam style+20 XP
A ball is thrown at \(12\ \text{m s}^{-1}\) from a point 5 m above the ground. Find its speed on reaching the ground.
Exam style+20 XP
A 2 kg particle slides 5 m down a rough plane (\(\sin\alpha = 0.6\), \(\mu = 0.25\)), starting at \(3\ \text{m s}^{-1}\). Find its final speed.
Exam style+20 XP
A 1000 kg car freewheels on a level road from \(25\ \text{m s}^{-1}\) to \(15\ \text{m s}^{-1}\) in 200 m. Find the constant resistance.
Exam style+20 XP
A cyclist (total 80 kg) rides 300 m up a road rising 15 m, slowing from \(8\) to \(5\ \text{m s}^{-1}\) against a resistance of 20 N. Find the work done by the cyclist.
Skill unlocked
Energy balanced.
This skill is now recorded on your profile. Keep the chain going.