An elastic spring of unstretched length $L$ and force constant $k$ is stretched by a small length $x$. It is further stretched by another small length $y$. Work done during the second stretching is

$A$ spring of force constant $10\, N/m$ has an initial stretch $0.20\, m$. In changing the stretch to $0.25\, m$,the increase in potential energy is about.....$joule$.

$A$ spring with a spring constant of $5 \times 10^3 \ N/m$ is initially in its natural state. It is stretched by $5 \ cm$. How much work is required to stretch it by another $5 \ cm$?

When equal forces are applied to two springs with spring constants $1500 \, N/m$ and $3000 \, N/m$,what is the ratio of their stored potential energies?

$A$ body of mass $0.15 \ kg$ moving with a velocity of $15 \ ms^{-1}$ comes to rest when it hits a spring that is fixed at the other end. If the force constant of the spring is $1500 \ Nm^{-1}$,then the compression in the spring is: (in $m$)

Statement-$1$ and Statement-$2$ are given. Two springs $S_1$ and $S_2$ have force constants $k_1$ and $k_2$ respectively,where $k_1 < k_2$. They are stretched by the same force $F$. It is found that the work done on spring $S_1$ is greater than the work done on spring $S_2$.
Statement-$1$: If stretched by the same force,the work done on $S_1$ is greater than the work done on $S_2$.
Statement-$2$: $k_1 < k_2$.