A flexible chain of length $L = 20 \sqrt{2} m$ and weight $W = 10 kg$ is initially placed at rest on a smooth frictionless wedge surface ABC. It is given a slight push on one-side so that sliding on that side. Calculate the speed of the chain, in ${ms}^{- 1}$ when its one end will leave the vertex of the wedge. Take $g = 10 {ms}^{- 2}$

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Detailed Solution

When chain leaves the vertex, then PE is decreased by the same factor by which KE is increased

$\sin (45^{\circ}) = \frac{h_{f}}{10 \sqrt{2}} \Rightarrow h_{f} = 10 m$

$Since, U_{f} = - {mgh}_{f} = - 10 \times 10 \times 10$

$\Rightarrow U_{f} = - 1000 J$

$Also, U_{i} = - {mgh}_{i} = - 10 \times 10 \times 5 \Rightarrow U_{i} = - 500 J$

$Since ΔU + ΔK = 0 \Rightarrow - 500 + \frac{1}{2} \times 10 \times v^{2} = 0$

$\Rightarrow v^{2} = 100 \Rightarrow v = 10 {ms}^{- 1}$

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