A block of mass m , lying on a smooth horizontal surface, is attached to a spring (of negligible mass) of…

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:

1. $\frac{F}{\sqrt{mk}}$
2. $\frac{2F}{\sqrt{mk}}$
3. $\frac{\pi F}{\sqrt{mk}}$
4. $\frac{F}{\pi \sqrt{mk}}$

Solution

When block attains maximum speed net force on $m$ is zero. $\therefore F=kx$

Or $x=\frac{F}{k}$

From work energy theorem

$-\frac{1}{2} k {\left(\frac{F}{k}\right)}^2+F\left(\frac{F}{k}\right)=\frac{1}{2}m{V}_{\mathrm{m}\mathrm{a}\mathrm{x}}^2$

$\frac{F^2}{2k}=\frac{1}{2}m{V}_{\mathrm{m}\mathrm{a}\mathrm{x}}^2$

$V_{\mathrm{m}\mathrm{a}\mathrm{x}}=\sqrt{\frac{F}{km}}$

Asked in: JEE Main 2019 (09 Jan Shift 1)

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