$A$ car of mass $1000 \,kg$ having a velocity of $10 \,ms^{-1}$ collides with a horizontally mounted spring. If the spring constant is $4000 \,Nm^{-1}$, then the maximum compression of the spring is: (in $\,m$)

The spring is initially in an unstretched condition. It is first stretched by a length $x$ and the work done is $W_1$. Then again it is stretched by a further length $x$ such that the total extension becomes $2x$. The work done in this second step is $W_2$. The value of $W_2$ is given by

$A$ block of mass $m$ is hung vertically from an elastic thread of force constant $\frac{2mg}{a}$. Initially,the thread is at its natural length and the block is allowed to fall freely. The work done on the block by the Earth when it passes through the equilibrium position will be:

An engine is attached to a wagon through a shock absorber of length $1.5 \, m$. The system with a total mass of $50,000 \, kg$ is moving with a speed of $36 \, km \, h^{-1}$ when the brakes are applied to bring it to rest. In the process of the system being brought to rest,the spring of the shock absorber gets compressed by $1.0 \, m$. If $90 \%$ of the energy of the wagon is lost due to friction,calculate the spring constant.

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

The potential energy of a long spring when stretched by $2\,cm$ is $U$. If the spring is stretched by $8\,cm$,the potential energy stored in it will be $.......\,U$.