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