Show that the law of conservation of mechanical energy is obeyed by pulling or compressing the block tied at the end of a spring.
In figure $(a)$, block is in equilibrium $\therefore x=0$
When a block of mass $m$ is pulled from its position $x=0$ upto $x_{m}$ and it released then it moves from $-x_{m}$ to $+x_{m}$. The total mechanical energy at any point $x=x$ it remains constant.
$\therefore \frac{1}{2} k x_{m}^{2}=\frac{1}{2} k x^{2}+\frac{1}{2} m v^{2}$
where $v$ is velocity of block at $x$.
At original position of block at $x=0$, the speed is maximum and hence kinetic energy is maximum.
$\therefore \frac{1}{2} m v_{m}^{2}=\frac{1}{2} k x_{m}^{2}$
where $v_{m}$ is maximum speed.
$\therefore v_{m}^{2}=\frac{k}{m} \cdot x_{m}^{2}$
$\therefore v_{m}=\sqrt{\frac{k}{m}} \cdot x_{m}$
Dimensional formula of $\sqrt{\frac{k}{m}}$ is [ $\mathrm{T}^{-2}$ ]
Hence dimensional formula of $v_{m}$ is same as the dimensional formula of $\sqrt{\frac{k}{m}} \cdot x_{m}$ Hence, kinetic energy is converted into potential energy and potential energy is converted into kinetic energy but total mechanical energy remains constant.
Explain the elastic potential energy of spring and obtain an expression for this energy.
A mass of $1\, kg$ is hanging from a spring of spring constant $1\, N/m$. If Saroj pulls the mass down by $2\,m$. The work done by Saroj is......$J$
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 energy of the wagon is lost due to friction, calculate the spring constant.
Find the maximum tension in the spring if initially spring at its natural length when block is released from rest.
If a long spring is stretched by $0.02\, m$, its potential energy is $U$. If the spring is stretched by $0.1\, m$ then its potential energy will be