In figure$\text{k}=100\text{N/m, m}=\text{1kg and F}=\text{10N}$ . A sharp blow by some external agent imparts a speed of $2\text{m/s}$ to the block towards left when it is at equilibrium position The potential energy of the spring when the block is at the right extreme is

Given ; $\text{K}=100\text{N/m,}\text{m}=\text{1}\text{kg,}\text{F}=10\text{N}$

a) In the equilibrium position compression

$\delta =\frac{\text{F}}{\text{K}}=\frac{10}{100}=0.1\text{m}$

$\delta =\text{10}\text{cm}$

b) The below imparts a speed of to the blocks towards left $\text{P}\text{.E}=\text{K}\text{.E}$

$\text{=}\frac{1}{2}\text{K}{\delta }^2+\frac{1}{2}\text{M}{\upsilon }^2=\frac{1}{2}\times 100{\left(0.1\right)}^2+\frac{1}{2}\times 1\times 4=2.5\text{J}$

c) The time period$\text{T}=2\pi \sqrt{\frac{\text{M}}{\text{k}}}=2\pi \sqrt{\frac{1}{100}}$

$\text{T}=\frac{\pi }{5}\sec$

d) Compression of the spring is$\left(x-\delta \right)$  since in SHM  the total energy remains constant

$\frac{1}{2}\text{K}{\left(x+\delta \right)}^2=\frac{1}{2}\text{K}{\delta }^2+\frac{1}{2}\text{M}{\upsilon }^2+\text{F}x$

$=2.5+10x\left[\because \frac{1}{2}\text{k}{\delta }^2+\frac{1}{2}\text{m}{\upsilon }^2=2.5\text{J}\right]$

$=\frac{1}{2}\left(100\right){\left(x+0.1\right)}^2=25+10x$

$\Rightarrow x=20\text{cm}$

e) Potential energy at the left extreme position

$\text{PE}=\frac{1}{2}\text{K}{\left(x+\text{A}\right)}^2$

$=\frac{1}{2}100{\left(0.2+0.1\right)}^2=50\times 0.09=4.5\text{J}$                        

f) P.E at the right extreme is given by

$\text{PE}=\frac{1}{2}\text{K}{\left(x+\delta \right)}^2-\text{F}\left(2x\right)$

$\because 2x=$ distance between two extremes

$=4.5-10\times 0.4$

$\text{PE}=0.5\text{J}$