Just over half length of the uniform chain of length L is hanging and the rest is placed over a rough table having coefficient of friction $μ = 0.5$ . If the chain is released from rest from this configuration then find speed of the chain at the moment it leaves the table completely.

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$\sqrt{\frac{3 gL}{8}}$

$\sqrt{\frac{3 gL}{4}}$

$\sqrt{\frac{gL}{8}}$

$\sqrt{\frac{5 gL}{8}}$

answer is D.

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Detailed Solution

Let say at some instant, the hanging length is y and length on table is (L-y).

$W_{f} = \int_{\frac{L}{2}}^{L} - 0.5 \frac{m (L - y) g}{L} dy = {- \frac{mg}{2 L} (Ly - \frac{y^{2}}{2}) |}_{\frac{L}{2}}^{L} = - \frac{mg}{2 L} ((L^{2} - \frac{L^{2}}{2}) - (\frac{L^{2}}{2} - \frac{L^{2}}{8})) = - \frac{mgL}{16} W_{g} = U_{i} - U_{f} = - \frac{m}{2} g \frac{L}{4} - (- \frac{mgL}{2}) = \frac{3 mgL}{8}$
Work energy theorem
$\Rightarrow W_{f} + W_{g} = \frac{1}{2} {mV}^{2} \Rightarrow - \frac{mgL}{16} + \frac{3 mgL}{8} = \frac{1}{2} {mV}^{2} \Rightarrow \frac{5 gL}{16} = \frac{V^{2}}{2} V = \sqrt{\frac{5 gL}{8}}$