This is a question I made up. I didn't really know a good way of checking if my approach was right. So.

Q: Consider a smooth surface on which a glass plate of mass $$M$$ is kept. If on this plate a wooden block of mass $$m$$ is given a horizontal velocity $$u$$, find the time taken by the block to come to rest relative to the glass plate given that the coefficient of kinetic and static friction between the block's surface and the glass plate is $$k$$.

Also predict how the block and the plate move after the block has come to rest relative to the glass plate.

Working from the frame of reference of the smooth surface, the only force on the block is the frictional force due to the plate. So its acceleration, $$a=-kg$$ and the acceleration of the table is $$b= \frac {km}{M} g.$$

We want to know the velocity of the table when the two are moving with the same velocity with respect to the smooth surface. Therefore, $$u+at=0+bt$$ Plugging in the values and solving we get $$t=\frac{Mu}{(M+m)kg}$$

Since there is no longer a tendency for relative motion, there is no more a frictional force between them. Hence, they continue to move with this new velocity.

Assumption- The glass plate is as long as required to ensure that the block does not slip off its end.

• so far I believe you are correct – Physicsapproval Dec 31 '16 at 3:28
• What is your question? I don't see one. The site policy is that "check my work" questions are not suitable here. You need to ask about a conceptual difficulty. – sammy gerbil Dec 31 '16 at 20:07

The force equations (from the friction force) for the block and the plate become \begin{align} \frac{dv_{\mathrm{block}}}{dt} = -kg \\ \frac{dv_{\mathrm{plate}}}{dt} = \frac{kmg}{M} \end{align} which together with the initial conditions $v_\mathrm{block}(0) = u$ and $v_\mathrm{plate}(0) = 0$ gives \begin{align} v_{\mathrm{block}} &= u - kgt \\ v_{\mathrm{plate}} &= \frac{kmgt}{M} \end{align} and setting them equal I get the same $t$ as you.
The only thing I can add is that the final velocity, which you can get by $$v_\mathrm{final} = \frac{kmgt}{M} = \frac{mu}{M+m}$$ can also easily be found by conservation of momentum: The initial momentum is $mu$ and the final momentum is $(M+m)v_\mathrm{final}$ and setting them equal directly gives you $v_\mathrm{final}$. It's also a good sanity check that your answer actually conserves momentum!
Also, it is customary to call the friction coefficients $\mu$ but it's really up to you.