Pushing a Wooden Crate

[caltzx]

1. The problem statement, all variables and given/known data
A wooden crate is pushed across a concrete floor at 5 m/s and released. It slides to a stop after moving a short distance. The same crate is filled until it weighs twice as much as it previously did and again slid across the floor at 5 m/s when released. The stopping distance for the crate will be...

a) 1/4 as far.

b) 1/2 as far.

c) the same distance.

d) twice as far.

e) four times as far.

3. The attempt at a solution
I thought this has something to do with kinetic energy. Since they both are released at 5 m/s but the latter has twice the mass, the second one should have twice the kinetic energy?

Therefore, I thought it would travel twice as far.

However, the answer given is: c) the same distance. Is this true? And if so, why?

Thanks. =)

[mgb_phys]

welcome to PF.
Yes Kinetic energy is = 1/2 m v^2 so with the same speed but twice the mass it will have twice the ke - correct.

But Friction force is = weight * coeff of friction.
And Energy is force * distance.

So how much extra friction does the second one have, how much extra energy does it have and so how far will it have to go to use up the extra energy.