Question

Consider the situation shown in figure (8-E8). Initially the spring is unstretched when the system is released from rest. Assuming no friction in the pulley, find the maximum elongation of the spring.
Figure

Answer 1

The maximum elongation of the unstretched spring when the system is released from rest (no friction in the pulley) is 2 mgh.

Explanation:

given data in the question  

Let body mass be m,

Assume the highest spring elongation is x.

As given No pulley friction, Then for the block , Potential energy reduction equals the potential energy accumulated in the spring.

Now, spring potential energy =  $\frac{1}{2} (k\times x^{2})$

and Potential energy transition = mgx  

Thus these two qualities are to be equated.

$\frac{1}{2} k\times x^{2} = mgx$

$k\times x^{2} =2\times mgx$

$x^{2} =\frac{2\times mgx}{k}$

$\frac{x^{2}}{x} =\frac{2\times mg}{k}$

$x =\frac{2\times mg}{k}$

Therefore the maximum elongation of unstretched spring is $x =\frac{2\times mg}{k}$ .

Answer 2

$\huge{\boxed{\mathcal\pink{\fcolorbox{red}{yellow}{Answer}}}}$

please refer the above attachment