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

Answer 1

The correct option is A $\frac{F}{\sqrt{mK}}$




Maximum speed is at mean position (equilibrium).
Let the extension in the spring is $x$.
So, the force acting on the block is
$F=Kx$
$x=\frac{F}{K}$
From work energy theorem
$W_F+W_{sp}=\Delta KE$

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

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

$v_{max}=\frac{F}{\sqrt{mK}}$

Final Answer: (c)