Q.
A block attached with a spring is kept on a smooth horizontal surface. Now the free end of the spring is pulled with a constant velocity $u$ horizontally. Then the maximum energy stored in the spring and block system during subsequent motion is
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Solution:
As $W_{\text {ext }}=\Delta( ME ) ; ME =$ Mechanical energy. Mechanical energy will keep on increasing up to the instant the $W_{\text {ext }}$ is positive, which will happen till there is no compression in the spring.
First the spring gets extended to a maximum and after which the extension decreases up to the natural length. After that there is a compression in the spring, results in a -ve external work (so as to move the end of spring at constant speed $u$ ). Hence, maximum energy stored is at the natural length and $ME _{\max }=(1 / 2) m v^{2}$
At the natural length $v=2 u$, since the block is moving at this instant at a speed $u$ with respect to the other end of the spring.
Hence, $ME _{\max }=(1 / 2) m(2 u)^{2}=2 m u^{2}$
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