Let us take the incline plane at angle thetaθ to the horizontal.
The force on the crate (mass mm) is the sum of the gravitational force - m g sin theta−mgsinθ (minus because it is down the slope) and the frictional force - m \mu−mμ (where \muμ is the coefficient of friction). However, the frictional force will change sign when the crate starts to move back down as it is opposite to the motion.
On the way up we have acceleration aa given by m a = - m g sin theta - m \muma=−mgsinθ−mμ
and we can cancel the mass mm to give
a = - g sin theta - \mua=−gsinθ−μ
On the way up the force (and therefore too the acceleration) is constant and so we have at time tt
v = u + a tv=u+at and s = u t + 1/2 a t^2 = (u + 1/2 a t)ts=ut+12at2=(u+12at)t.
At the top of the motion v=0v=0 and so t=-u/at=−ua, from which s = -u^2/(2a)s=−u22a (recall aa is negative).
Now we have to come back down; the acceleration will be different, say a' = - g sin theta + \mu, as the direction of motion has reversed. For simplicity, we reset the clock to t=0 (with therefore a new u=0), and so now
v = a' t and s = s_0 + 1/2 a' t^2,
where s_0=-u^2/(2a) is just the new starting point (the top of the motion).
The return point is just s=0 and so 0 = s_0 + 1/2 a' t^2, which gives
t = \sqrt(u^2/(aa')) and so v = u\sqrt((a')/a) = u\sqrt((g sin theta - \mu)/(g sin theta + \mu))** .
