Formula to solve this?

Question

the driver of an 800kg car decides to double the speed from 20m/s to 40 m/s. What effect would this have on the amount of work required to stop the car: that is, on the kinetic energy of the car?

Answer 1

KE = 1/2mv^2

If v doubles, the kinetic energy (KE) quadruples. That means you'd need 4 times the energy to stop the car.

Take it from an engineering student

Answer 2

KE=0.5mv^2

Work=KE

W(before)=0.5*800*20^2=160000

W(after)=0.5*800*40^2=64000

Therefore work require is increased 4 fold.

One could also do this without calculation by arguing that, using the above eqatuion, as the speed doubles, the KE increases 4 fold due to the v^2 term. This is slightly lazy in my opinion, it's something an engineering student would do! Go physicists! :op

Answer 3

Kinetic Energy is 1/2mv^2

Work(before) = .5(800kg)(20m/s)^2

Work(after) = .5(800kg)(40m/s)^2