Question

A spring of stiffness K is kept compressed by applying horizontal force on a block of mass m by a length $x_0=\left(\frac{mg}{k}\right)$ as shown in the figure. If the force F is withdrawn suddenly, the block oscillates and finally stops.
In consequence, fractional loss of energy is equal to $50\%$ of the initial potential energy stored in the spring.
The coefficient of friction between block and ground is found as $[x{]}^{\frac{-1}{2}}$ ; what is x? ___

Answer 1

Since $50\%$ of energy lost due to friction hence energy remains associated with springs =$\frac{m^2{g}^2}{4k}$
$\Rightarrow \frac{1}{2}k{x}^2=\frac{m^2{g}^2}{4k^2};x=\frac{mg}{\sqrt{2k}};kx=\mu mg$
$\therefore x=2$