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A plank of length $L$ and mass $M$ is placed on the ground such that there is no friction between the plank and the ground. A small block with mass $m$ is placed on one side of the plank. Friction between block and plank is $\mu$. At point $t=0$ block has some velocity $v_0$. What is the minimal $v_0$ such that block can travel to the end of the plank?

There is an image for easier understanding.

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There are two ways to solve this question. First method if to look relative to the plank, so we must add inertial force $F_{in}$ to block. Second method is to look relative to the ground. I will explain the second method, so for now forget about $F_{in}$.

From the image, we can create four equations from force vector addition. $$ma_1=-T\\0=N-mg\\Ma_2=T\\0=N_P-N-Mg$$ Here $N_P$ is the force ground react to the plank, but it is not important. From well-known relation $T=\mu N$ we have $$a_1=-\mu g\\a_2=-a_1\frac m M$$ Because both $a1$ and $a2$ are constant, we can apply equation $$v_1^2=v_0^2+2a_1s_1$$ where $v_1$ is velocity of the block at some point and $s_1$ is the distance block traveled. Because $v_0$ sould be minimal, the block stops at the end of the plank, so $v_1=0$ and $$s_1=-\frac{v_0^2}{2a_1}$$ Similary, from $$v_1=v_0+a_1t$$ we find that time to required for traveling to the end of the plank is $$t=-\frac{v_0}{a_1}$$ This is all we need for the block. Now, for the plank we need to find the distance it traveled for time $t$ $$s_2=v_{0_P}t+\frac12a_2t^2\\s_2=-\frac mM\cdot\frac{v_0^2}{2a_1}$$ where $v_{0_P}$ is velocity of the plank at $t=0$ which is $0$. Finally, we have the relation between distance traveled between the block and the plank $$s_1=s_2+L$$ It gives us $$v_0=\sqrt{\frac{2\mu gLM}{M-m}}$$ which is, for some reason, incorrect solution. I simply cannot see where I did a mistake, but I am sure I missed something obvious. Did I simply misscalculated something it these equations?

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The work-energy theorem and conservation of momentum should help in solving this problem.

Initially the small block is moving with speed $v_0$ relative to the ground while the plank is at rest. Finally the small block and plank are moving together with common speed $V$. While momentum is conserved, some KE is lost as the small block slides against friction from one end of the plank to the other. Assuming the length of the small block is much smaller than $L$, then the work done against friction is the friction force $\mu mg$ times the distance $L$, and this equals the KE lost. As you have found (see comment below) the result is
$v_0 = \sqrt{2\mu gL(1+\frac{m}{M})}$.


The mistake in your calculation is to assume that the final velocity of the small block is zero relative to the ground : $v_1=0$. Instead, this velocity is zero relative to the plank.

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  • $\begingroup$ Which means that final solution is $v_0=\sqrt{2\mu gL\left(1+\frac mM\right)}$, or am I wrong? Also, it doesn't explaint where is a mistake in my solution. $\endgroup$ – sbisit Oct 14 '16 at 22:45
  • $\begingroup$ Yes, that is the solution I get too. Probably your mistake is that you assume the final velocity of the small block relative to the ground is zero, which is not correct. $\endgroup$ – sammy gerbil Oct 14 '16 at 22:58

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