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Kinetic energy
The energy possessed by a body by virtue of its motion called the kinetic energy of the
body. A rocket traveling to the Moon has kinetic energy as does a snail crawling along a wall.
It is the kinetic energy of objects that makes them difficult to stop and the kinetic energy of
the air in a hurricane that causes severe damage to the countryside.
The kinetic energy of an object depends on two things:
(a) the mass of the object (m)
(b) its speed (v)
The formula for kinetic energy of an object of mass m travelling at velocity v is:
Kinetic energy (k.e) = ½ mv2
Proof of the formula for the kinetic energy of an object
Suppose that a body of mass m initially at rest is accelerated by constant force F so that it
reaches a velocity v after travelling a distance of s metres. (Figure 1).
velocity (v)
velocity (v) = 0
Mass
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Accelerating distance (s)
(m)
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Figure 1
The kinetic energy gained by the body is equal to Fs, and this must therefore be the work
done in giving it a velocity v.
However we know that F = ma and v2 = u2 + 2as where u and v are the initial and final
velocities of the object. In this case u = 0 and so we have:
v2= 2as and so ½v2 = as
Therefore:
kinetic energy = Fs = mas = ½ mv2
Kinetic energy = ½ mv2
The kinetic energy of an object is measured in Joules (J), kilojoules (kJ) or Megajoules (MJ).
Estimate of the kinetic energies of various objects:
speeding car
sprinter
oil tanker
air molecule
650 kJ
5 kJ
5000 MJ
2.5x10-21 J
1
Kinetic energy changes
It is important to understand the correct way to calculate changes in the kinetic energy of an
object.
For example suppose we want to find the increase in the kinetiec energy of an 8kg ball when
its velocity is increased from 3 ms-1 to 4 ms-1. The correct way is as follows:
Kinetic energy increase (ke) = ½ x8x[42 – 32] = 4x[16-9] = 4x7= 28 J
and NOT
Kinetic energy increase = ½ x8x[4-3]2 = 4 J
Example problem
A lorry of mass 6000 kg travels along a level road a 30 ms -1. The brakes are then
applied and the lorry stops in 70 m. Calculate:
(a) the kinetic energy of the lorry before braking
(b) the braking force
(a) kinetic energy = ½ mv2 = ½ x 6000 x 302 = 2 700 000 J = 2.7 MJ
(b) braking force = Energy changed/ Braking distance = 2.7x106/70 = 38.6 kN
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