A 2.4 kg piece of wood slides on the surface shown in the figure . The curved sides are perfectly smooth, but the rough horizontal bottom is 31m long and has a kinetic friction coefficient of 0.27 with the wood. The piece of wood starts from rest 4.0m above the rough bottom.
a) Where will this wood eventually come to rest?
b) For the motion from the initial release until the piece of wood comes to rest, what is the total amount of work done by friction?
E_k = (1/2)mv^2
E_f = mg(u_k)d
E_g = mgh
The Attempt at a Solution
I think I have an idea for part a but I'm not sure. Would this be correct?
Initial Energy = Final Energy
Potential Gravitational Energy = Friction Energy
mgh = Ugdm
(2.4kg)(9.8m/s^2)(4.0m) = 0.27(9.8m/s^2)(2.4kg)d
And then I figured that I would solved for d...
Would that be correct?
And for part b would I use the work kinetic energy theorem even though it has friction?