Two cars have the same mass, but one is moving three times faster than the other. How much more work will be needed to stop the faster car?

Answer 1

KE = (1/2) m v^2
You must do enough work (Force times braking distance) to equal the original Kinetic energy.
W1 = (1/2) m v^2
W2 = (1/2) m (3v)^2 = (1/2) m v^2 * 9
so
nine times as much

Answer 2

I guess NINE TIMES AS MUCH