Question

A block of mass m sliding on a smooth horizontal surface with a velocity \[\vec{\nu}\] meets a long horizontal spring fixed at one end and with spring constant k, as shown in following figure following. Find the maximum compression of the spring. Will the velocity of the block be the same as  \[\vec{\nu}\]  when it comes back to the original position shown?

Answer 1

Let the compression in the spring be x.
(a) Applying the law of conservation of energy,
maximum compression in the spring will be produced when the block comes to rest .
so change in kinetic energy of the block due to change in its velocity from u m/s to 0 will be equal to the gain in potential energy of the spring.
change in kinetic energy of the block= \[\frac{1}{2}\text{mv}^2 - \frac{1}{2}\text{m}(0 )^2 = \frac{1}{2}\text{mv}^2\]

gain in the potential energy of spring= \[\frac{1}{2}\text{kx}^2\]

\[\frac{1}{2}\text{m} \nu^2 = \frac{1}{2}\text{kx}^2 \]

\[ \Rightarrow \text{x}^2 = \frac{\text{m} \nu^2}{\text{k}}\]

\[\text{x} = \nu\sqrt{\left( \frac{\text{m}}{\text{k}} \right)}\]

(b) No. The velocity of the block will not be same when it comes back to the original position. It will be in the opposite direction and the magnitude will be the same if we neglect all losses due friction and spring to be perfectly elastic.