question_answer

If you are pulling one end of an ideal spring of mass m with velocity v, while keeping other end fixed to a wall, then kinetic energy of the spring at this instant is

A) \[\frac{1}{2}m{{v}^{2}}\]

B) \[\frac{1}{4}m{{v}^{2}}\]

C) \[\frac{1}{6}m{{v}^{2}}\]

D) \[\frac{1}{8}m{{v}^{2}}\]

Correct Answer: C

Solution :

Mass dm of spring at a distance x moves with velocity is \[{{v}_{x}}=\frac{x}{l}\cdot v\]

Kinetic energy of mass dm is \[dK=\frac{1}{2}(dm)v_{x}^{2}\]\[\Rightarrow \]\[dK=\frac{1}{2}\cdot \frac{{{v}^{2}}{{x}^{2}}}{{{l}^{2}}}\cdot dm\]

Kinetic energy of complete spring is \[K=\int{dK=\frac{1}{2}\cdot \frac{{{v}^{2}}}{{{l}^{2}}}\cdot \int\limits_{0}^{l}{{{x}^{2}}dm}}\]

Here, \[dm=\frac{m}{l}dx\]

\[\therefore \]\[K=\frac{1}{2}\cdot \frac{{{v}^{2}}}{{{l}^{2}}}\times \frac{m}{l}\int\limits_{0}^{l}{{{x}^{2}}dx=\frac{1}{6}m{{v}^{2}}}\]