Plzz solve this question no 33

Dear student
The potential energy of a spring is the energy stored by a spring due to its compression.
According to Hooke's law, the restoring force produced in a spring is given by,
F=-KX
Where, k=Constant
x=Displacement produced
Now, the potential energy would be equal to the work done on the restoring system,
Therefore, P.E=W=F.dx
$$\begin{gathered} \mathrm{On} \mathrm{integration}, \mathrm{we} \mathrm{get} \\ \mathrm{P}.\mathrm{E}.=-\frac{1}{2}{\mathrm{kx}}^2 \end{gathered}$$
If the work done by a force does not depends upon the initial and final point but depends upon the path taken, it is called a non conservative force.
Spring force is non conservative force because work done depends upon the actual path. 
$$\begin{gathered} \mathrm{For} \mathrm{small} \mathrm{displacement}, \mathrm{spring} \mathrm{force} \mathrm{F}=-\mathrm{kx} \\ \mathrm{The} \mathrm{elastic} \mathrm{potential} \mathrm{energy} \mathrm{of} \mathrm{the} \mathrm{spring}=\mathrm{Work} \mathrm{done} \mathrm{on} \mathrm{it} \mathrm{to} \mathrm{deform} \mathrm{it} \mathrm{an} \mathrm{amount} \mathrm{x} \\ \mathrm{Thus}, \mathrm{V}=-{\int }_0^{\mathrm{x}}-\mathrm{kx} \mathrm{dx}=\frac{1}{2}{\mathrm{kx}}^2 \\ \mathrm{Here} \mathrm{if} \mathrm{the} \mathrm{deformation} \mathrm{increases} \mathrm{from} {\mathrm{x}}_1 \mathrm{to} {\mathrm{x}}_2 \mathrm{during} \mathrm{the} \mathrm{motion}, \mathrm{then} \mathrm{the} \mathrm{change} \mathrm{in} \mathrm{potential} \mathrm{energy} \mathrm{of} \mathrm{the} \mathrm{spring} \mathrm{will} \mathrm{be} \mathrm{the} \mathrm{difference} \mathrm{between} \mathrm{its} \mathrm{final} \mathrm{and} \mathrm{initial} \mathrm{values}. \\ ∆\mathrm{V}=\frac{1}{2}\mathrm{k}({{\mathrm{x}}_2}^2-{{\mathrm{x}}_1}^2) \end{gathered}$$
Work done is equal to change in potential energy thus it is conservative force.
kindly post another question in different thread.
Regards