A small disc of mass m slides down a smooth hill of height h without initial velocity and gets onto a plank of mass M lying on a smooth horizontal plane at the base of hill figure. Due to friction between the disc and the plank, disc slows down and finally moves as one piece with the plank. Find out total work performed by the friction forces in this process.

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$W_{f r} = - m g h [\frac{M}{M + m}]$

$W_{f r} = - g h [\frac{M}{M + m}]$

$W_{f r} = - m h [\frac{M}{M + m}]$

$W_{f r} = - m g h [\frac{M}{m}]$

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Detailed Solution

When the disc slides down and comes onto the plank, then

$m g h = \frac{1}{2} m v^{2} v = \sqrt{(2 g h)}$

Let $v_{1}$ be the common velocity of both, the disc and plank when they move together. From law of conservation of linear momentum,

$m v = (M + m) v_{1} v_{1} = \frac{m v}{(M + m)}$

Now, change in $KE = (K)_{f} - (K)_{i} = {work done}_{triction}$

$∴ \frac{1}{2} (M + m) v_{1}^{2} - \frac{1}{2} m v^{2} = {work done}_{friction}$

$W_{fr} = \frac{1}{2} (M + m) {[\frac{m v}{M + m}]}^{2} - \frac{1}{2} m v^{2}$

$= \frac{1}{2} m v^{2} [\frac{m}{M + m} - 1]$

$\frac{1}{2} m v^{2} = m g h W_{fr} = - m g h [\frac{M}{M + m}]$

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