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Question

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:

![Block-spring on smooth surface with constant force F](https://www.edvaya.com/jee/physics/q33-block-spring-constant-force-2019.png) A). $\dfrac{2F}{\sqrt{mk}}$ B). $\dfrac{F}{\pi\sqrt{mk}}$ C). $\dfrac{\pi F}{\sqrt{mk}}$ D). $\dfrac{F}{\sqrt{mk}}$

Edvaya · Accepted Answer

The correct answer is $\dfrac{F}{\sqrt{mk}}$. Maximum speed occurs when net force on block = 0, i.e., when spring force equals applied force:
$kx = F \Rightarrow x = \dfrac{F}{k}$
Using work-energy theorem from equilibrium to this point:
$W_{net} = \Delta KE$
$Fx - \dfrac{1}{2}kx^2 = \dfrac{1}{2}mv_{max}^2$
$F \cdot \dfrac{F}{k} - \dfrac{1}{2}k\dfrac{F^2}{k^2} = \dfrac{1}{2}mv_{max}^2$
$\dfrac{F^2}{k} - \dfrac{F^2}{2k} = \dfrac{1}{2}mv_{max}^2$
$\dfrac{F^2}{2k} = \dfrac{1}{2}mv_{max}^2$
$v_{max} = \dfrac{F}{\sqrt{mk}}$

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