A man stretches a horizontal spring of spring constant k attached to the wall of a railway carriage with uniform velocity‘v’ relative to carriage as shown in figure. Carriage is moving in same direction as stretching with uniform speed ‘u’. In ground's frame of reference

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Work done by man on spring in time t is $\frac{1}{2} k v (v + u) t^{2}$

Work done by man on spring in time t is $\frac{1}{2} k v (v - u) t^{2}$

Work done by trolley on spring in time t is $- \frac{1}{2} k u (v - u) t^{2}$

Work done by trolley on spring in time t is $- \frac{1}{2} k u v t^{2}$

answer is B, D.

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Detailed Solution

At time t elongation in spring = vt
Spring force = kvt
$\begin{array}{l} ∴ work done by the man on spring = \int_{0}^{t} (kvt) (u + v) dt = \frac{1}{2} kv (u + v) t^{2} \\ ∴ work done by trolley on spring = \int_{0}^{t} (kvt) (udt) (- 1) = - \frac{1}{2} {kuvt}^{2} \end{array}$