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Unlock This CourseYou now know three equations that each count energy in joules:
Remember, energy is conserved. It cannot be created or destroyed, only transferred. So the energy that leaves one store must arrive in another. This means you can put one equation equal to another.
| Situation | Energy statement |
|---|---|
| An object falls (air resistance ignored) | GPE lost = KE gained |
| An object is thrown upwards (air resistance ignored) | KE lost = GPE gained |
| A force speeds an object up (friction ignored) | work done = KE gained |
| Brakes or friction slow an object down | work done = KE lost |
There is no new equation to learn here, but none of the three above is given on the formulae page.
When an object falls, energy is transferred from its gravitational store to its kinetic store. If we ignore air resistance, all of the GPE lost becomes KE:
m × g × h = ½ × m × v2
The mass m is on both sides, so it cancels. That leaves g × h = ½ × v2, which gives:
v2 = 2 × g × h
So the speed at the bottom depends only on the height fallen, not on the mass.
A diver of mass 50 kg steps off a board 3.2 m above the water. Calculate her speed as she reaches the water. Ignore air resistance.
GPE lost = m × g × h = 50 × 10 × 3.2 = 1600 J
KE gained = 1600 J
v2 = (2 × KE) ÷ m = (2 × 1600) ÷ 50 = 64
v = √64 = 8 m/s
Short way: v2 = 2 × g × h = 2 × 10 × 3.2 = 64, so v = 8 m/s. An 80 kg diver loses more GPE (2560 J), but v2 = (2 × 2560) ÷ 80 = 64 again.
Thrown upwards, the transfer runs the other way. At the very top the speed is zero for an instant, so all of the KE the object started with has become GPE.
A tennis ball is thrown straight up at 6 m/s. Calculate the greatest height it reaches above the hand. Ignore air resistance.
KE lost = GPE gained
½ × m × v2 = m × g × h, and the mass cancels
h = v2 ÷ (2 × g)
h = 62 ÷ (2 × 10) = 36 ÷ 20
h = 1.8 m
At the top of the first hill the train has its greatest GPE and is almost at rest. As it drops, GPE is transferred to KE, so it is fastest at the lowest point. Climbing the next hill, KE is transferred back to GPE and it slows down. For a 20 m drop: v2 = 2 × 10 × 20 = 400, so v = 20 m/s.
A pendulum is a small mass (the bob) swinging on a string. At the highest point on each side the bob stops for an instant: greatest GPE, no KE. At the lowest point, in the middle, it is fastest: greatest KE, least GPE. If the bob is released 0.2 m above its lowest point: v2 = 2 × 10 × 0.2 = 4, so v = 2 m/s.
When a force speeds an object up on level ground, with friction ignored, all of the work done ends up in the kinetic store:
work done by the force = KE gained
A sledge and child have a total mass of 50 kg. They start at rest on smooth, level ice. A parent pushes with a force of 90 N for 10 m. Calculate the final speed. Ignore friction.
work done = F × d = 90 × 10 = 900 J
KE gained = 900 J
v2 = (2 × KE) ÷ m = (2 × 900) ÷ 50 = 36
v = √36 = 6 m/s
Stopping is the reverse. To stop an object, the brakes or friction must do work equal to all of its kinetic energy:
work done by brakes or friction = KE lost
A minibus of mass 2000 kg travels at 15 m/s. The brakes give a braking force of 7500 N. Calculate the braking distance.
KE = ½ × m × v2 = ½ × 2000 × 152 = ½ × 2000 × 225 = 225 000 J
work done by brakes = 225 000 J
d = W ÷ F = 225 000 ÷ 7500
d = 30 m
The same method finds a braking or friction force. Work out the KE lost, then use F = W ÷ d with the stopping distance.
Real objects never quite reach the speeds worked out above. Friction and air resistance do work on the moving object. This transfers some energy to the thermal stores of the object and its surroundings, which warm up slightly. The energy has not been destroyed. So:
GPE lost = KE gained + energy transferred to thermal stores
A roller coaster train of mass 1000 kg drops 20 m and reaches 18 m/s, not 20 m/s.
GPE lost = 1000 × 10 × 20 = 200 000 J
KE gained = ½ × 1000 × 182 = ½ × 1000 × 324 = 162 000 J
energy transferred to thermal stores = 200 000 − 162 000 = 38 000 J
This is the work done against friction and air resistance. It heats the wheels, the track and the air.
A bouncing ball shows the same thing. As it falls, GPE becomes KE. When it hits the floor it squashes, and the energy moves into its elastic store. Then it springs back into shape, the energy returns to the kinetic store, and the ball rises, turning KE back into GPE. During each bounce some energy is transferred to the thermal stores of the ball and the floor. So the ball leaves the floor with less KE than it arrived with, and each bounce is lower than the last.
For example, a ball of mass 0.05 kg dropped from 2.0 m starts with 0.05 × 10 × 2.0 = 1.0 J of GPE. If it rebounds to 1.2 m it has 0.6 J, so 0.4 J has been transferred to thermal stores. A pendulum left swinging slowly stops for the same reason.
Key terms:
You can investigate the factors that affect the rebound height of a bouncing ball. Hold a metre rule upright against a wall, drop the ball from a measured height and read the rebound height at the bottom of the ball. Repeat and take a mean. Change one factor at a time, such as the drop height, the type of ball, the surface it lands on or the temperature of the ball.
A child of mass 40 kg sits at the top of a water slide. The top of the slide is 7.2 m above the bottom. The child starts from rest.
(a) Calculate the gravitational potential energy the child loses in sliding to the bottom. [2 marks]
(b) Calculate the speed the child would have at the bottom if there were no friction or air resistance. [3 marks]
(c) The child's actual speed at the bottom is 9 m/s. Calculate the energy transferred to thermal stores during the slide. [3 marks]
(d) At the bottom, the child moves along a level pool of water and stops in a distance of 4.5 m. Calculate the average resistive force on the child. [2 marks]
(a) GPE = m × g × h = 40 × 10 × 7.2 (1) = 2880 J (1)
(b) KE gained = GPE lost = 2880 J, so v2 = (2 × KE) ÷ m = (2 × 2880) ÷ 40 (1) = 144 (1), so v = √144 = 12 m/s (1)
(c) KE = ½ × 40 × 92 (1) = ½ × 40 × 81 = 1620 J (1). Energy transferred to thermal stores = 2880 − 1620 = 1260 J (1)
(d) work done by the resistive force = KE lost = 1620 J, so F = W ÷ d = 1620 ÷ 4.5 (1) = 360 N (1)
Start each part by writing the energy statement in words, such as GPE lost = KE gained. It tells you which two equations to join. In part (d), use the 1620 J the child really has, not 2880 J.