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The upper end of the uniform chain of linear mass density 2 kg/m is lowered with a speed 1 m/s as shown in diagram. The normal reaction offered by the ground after the chain fall through a distance 5 cm is given by k newton. Find the value of k. $g = 10 m / \sec^{2}$

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answer is 3.

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Detailed Solution

Each falling link collides with the fallen links with a relative velocity

${\bar{V}}_{i m p} = - v j,$

${\bar{f}}_{i m p} = {\bar{V}}_{r e l} \frac{d m^{1}}{d t}, {\bar{f}}_{i m p} = - μ V^{2} j$

From equation on heap

$\begin{array}{l} N - m' g + f_{i m p} = 0 \\ N = μ (V^{2} + g y) = 2 (1^{2} + 10 \times 0.05) = 2 (1 + 0.5) = 3 N \end{array}$

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