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

• A. $\frac{\pi F}{\sqrt{mk}}$
• B. $\frac{F}{\sqrt{mk}}$
• C. $\frac{2F}{\sqrt{mk}}$
• D. $\frac{F}{\pi \sqrt{mk}}$

Answer 1

The correct option is B $\frac{F}{\sqrt{mk}}$

According to the work-energy theorem,
net work done = change in the kinetic energy
Here, net work done = work done due to external force $\left(W_{ext}\right)$ + work done due to the spring $\left(W_{spr}\right)$
As, $W_{ext}=F.x$
and $W_{spr}=\frac{-1}{2}k{x}^2$
$\Rightarrow \Delta KE=F.x+(-\frac{1}{2}k{x}^2)$
$(KE{)}_f-(KE{)}_i=F.x-\frac{1}{2}k{x}^2$
Maximum speed is at equilibrium position, when $F=kx\Rightarrow x=\frac{F}{k}$
Thus,
$\frac{1}{2}m{v}_{max}^2-\frac{1}{2}m(0{)}^2=F.\left(\frac{F}{k}\right)-\frac{1}{2}k{\left(\frac{F}{k}\right)}^2$
$\Rightarrow \frac{1}{2}m{v}_{max}^2=\frac{F^2}{k}-\frac{F^2}{2k}=\frac{F^2}{2k}$
or $v_{max}^2=\frac{F^2}{km}$
$\Rightarrow v_{max}=\frac{F}{\sqrt{mk}}$