Q 11-13-135JEE MainJEE Main 2019 (9 Jan, Shift 1)Medium
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
Answer: (A) $\dfrac{F}{\sqrt{mk}}$
With the constant force $F$, the new equilibrium position is at an extension $x_0 = F/k$. The block starts from rest at $x = 0$, so it performs SHM about $x_0$ with amplitude $A = F/k$ and $\omega = \sqrt{k/m}$.
The maximum speed is at the new equilibrium position:
$$v_{\max} = A\omega = \frac{F}{k}\sqrt{\frac{k}{m}} = \frac{F}{\sqrt{mk}}$$
Solution by Sreeraj P, M.Sc Physics