Set 3 Problem number 9

A mass of 60 kilograms is moving at 7 meters/second. Find its kinetic energy, its kinetic energy if its speed doubled, and the factor by which kinetic energy increases when speed is doubled.

Solution

When v is 7 m/s and m is 60 Kg, we have

KE= .5( 60 Kg)( 7 m/s)^2 = .5( 60 Kg)( 7 m/s)^2 = 1470 Joules.

The same procedure, using 14 m/s, yields kinetic energy 5880 Joules.

The factor by which the kinetic energy increases is the ratio 5880/ 1470 = 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.