Question

A spring of unstretched length $l$ has a mass $m$ with one end fixed to a rigid support. Assuming spring to be made of a uniform wire, the kinetic energy possessed by it if its free end is pulled with uniform velocity $v$ is

• A. $\frac{1}{2}m{v}^2$
• B. $m{v}^2$
• C. $\frac{1}{3}m{v}^2$
• D. $\frac{1}{6}m{v}^2$

Answer 1

The correct option is D $\frac{1}{6}m{v}^2$

Let $\omega =\frac{m}{l}$(linear mass density)

Velocity of dx element and a distance x from rigid support is $v^{\prime }=\frac{vx}{l}$

Now mass of thickness (dx)$=\omega dx=dm$

KE of particle of thickness$=\left(dx\right)=\frac{1}{2}dm{v}^{\prime 2}$

$KE={\int }_0^l\frac{1}{2}\omega dx{\left(\frac{vx}{L}\right)}^2=\frac{1}{2}\frac{m}{L}\frac{v^2}{L^2}{\int }_0^L{x}^{2}dx=\frac{m{v}^2}{2L^3}\times \stackrel{L}{\underset{0}{\left[\frac{x^3}{3}\right]}}=\frac{m{v}^2}{6}$