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JEE Main 2019 (9 Jan, Shift 1) · Spring & Elastic Energy

Spring & Elastic Energy question from JEE Main 2019 (9 Jan, Shift 1)

Q19JEE Main 2019 · 9 JAN · SHIFT 1Difficult

Spring & Elastic Energy

A block of mass m, lying on a smooth horizontal surface, is attached to a spring (of negligible mass) of spring constant k. The other end of the spring is fixed, as shown in the figure. The block is initially at rest in its equilibrium position. If now the block is pulled with a constant force F, the maximum speed of the block is:

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Answer: (d)

Speed is maximum where net force is zero: $kx = F \Rightarrow x = F/k$ .
Work-energy: $Fx - \frac{1}{2}kx^2 = \frac{1}{2}mv^2$
$\frac{F^2}{k} - \frac{F^2}{2k} = \frac{F^2}{2k} = \frac{1}{2}mv^2 \Rightarrow v = \frac{F}{\sqrt{mk}}$
So the answer is (d) $\frac{F}{\sqrt{mk}}$ .

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