$A$ spring has a spring constant $200 \,N/m$. If it is stretched by $1 \,cm$, then the potential energy stored in it is: (in $\,J$)

$A$ block of mass $m$ slides down an inclined plane as shown in the figure and hits a spring at the bottom,causing it to compress. If the spring constant is $K$,what is the maximum compression of the spring?

The force constants of two springs are $K_1$ and $K_2$. Both are stretched until their elastic potential energies are equal. If the stretching forces applied are $F_1$ and $F_2$,then the ratio $F_1:F_2$ is:

Two springs of spring constants $K$ and $2K$ are stretched by the same force. If $W_{1}$ and $W_{2}$ are the energies stored in them respectively,then:

$A$ spring with spring constant $k$ when stretched through $1 \, cm$,the potential energy is $U$. If it is stretched by $4 \, cm$,the potential energy will be (in $U$)

Show that the law of conservation of mechanical energy is obeyed by pulling or compressing the block tied at the end of a spring.