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

Answer 1

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

Maximum speed is at mean position or equilibrium .

At extreme position
$F=kx\Rightarrow x=\frac{F}{k}$

From work-energy theorem,
$W_F+W_{sp}=\Delta KE$

$F\left(x\right)-\frac{1}{2}k{x}^2=\frac{1}{2}m{v}^2-0$

$F\left(\frac{F}{k}\right)-\frac{1}{2}k{\left(\frac{F}{k}\right)}^2=\frac{1}{2}m{v}^2$

$\Rightarrow \frac{1}{2}\frac{F^2}{k}=\frac{1}{2}m{v}^2$

$\therefore v_{max}=\frac{F}{\sqrt{mk}}$

Hence, option $\left(\text{D}\right)$ is correct.