Physics help!?

Question

A small block slides along a track from one level to a higher level, by moving through an intermediate valley (see Figure). The track is frictionless until the block reaches the higher level. There a frictional force stops the block in a distance d. Assume that the block's initial speed is 11 m/s, the height difference h is 3.5 m, and muk is 0.66. Find the distance d that the block travels on the higher level before stopping.

Answer 1

This is solved by energy conservation. Find the block's total energy when it reaches the frictional section: The energy at that point has come from two sources: the initial kinetic energy from its intial velocity, which is .5*m*v0^2 and the change in potential energy due to the elevation difference between the two levels, which is m*g*(L1 - L2) (note that L2 is greater than L1 so that there is loss of total energy due to potential energy change). Therefore the block's energy on arrival at the higher level is .5*m*v0^2 + m*g*(L1 - L2); the energy lost to friction is F*d, where F is the force of friction and d is the travel distance (this eq is true if F is constant). The force from friction is the weight of the mass times the coefficient of friction, or m*g*µk equating this to the total energy gives

d*m*g*µk = .5*m*v0^2 + m*g*(L1 - L2)

Notice that the mass cancels out (which is why it wasn't specified) leaving

d*g*µk = .5*v0^2 + g*(L1 - L2)

d = [.5*v0^2 + g*(L1 - L2)]/g*µk

g = 9.8m/sec^2