Set 3 Problem number 9

The kinetic energy of an object is equal to the work required of the net force to accelerate it from rest to its present velocity. It is easily enough shown that this work is equal to .5 m v^2, where m is the mass and v the velocity of the object. This quantity is independent of the acceleration. We therefore say that the kinetic energy of the object is .5 m v^2.

If an object of mass 53 kg is moving at 14 m/s, what is its KE?

If its velocity doubles, what is its KE.
By what factor does its KE increase when its velocity doubles?

Solution

When v is 14 m/s and m is 53 Kg, we have

KE= .5( 53 Kg)( 14 m/s)^2 = .5( 53 Kg)( 14 m/s)^2 = 5194 Joules.

The same procedure, using 28 m/s, yields kinetic energy 20776 Joules.

The factor by which the kinetic energy increases is the ratio 20776/ 5194 = 4 of these results.

Generalized Solution

At velocity v, the kinetic energy of a mass m is KE1 = .5 m v^2.

At velocity v ' = 2v, the kinetic energy is

KE2 = .5 m (v ' )^2 = .5 m (2v) ^ 2 = .5 m (4 v^2).

The ratio of kinetic energies is therefore

KE2 / KE1 = .5 m (4 v^2) / (.5 m v^2) = 4.

More generally if velocity increases by factor c:

From v to v' = cv, the kinetic energy will increase from KE1 = .5 m v^2 to KE2 = .5 m (cv)^2 = .5 m (c^2 v^2), and the ratio of kinetic energies will be KE2 / KE1 = c^2.

Kinetic energy thus increases by a factor equal to the square of the velocity increase.

As seen above doubling velocity doubles kinetic energy.
If velocity is tripled kinetic energy will increase by factor 3^2 = 9.
If velocity increases to 10 times its original value kinetic energy will increase by factor 100.

This, incidentally, is the reason why it is so much worse to run your car into a solid object at 60 mph than at 30 mph: the KE is four times as great at 60 mph than at 30 mph.
It also takes 4 times as far to stop a car from 60 mph than from 30 mph.