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 ν is:

# 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 $\mathrm{\nu }$ is:

1. A

$\frac{1}{2}{\mathrm{m\nu }}^{2}$

2. B

$\frac{1}{3}{\mathrm{m\nu }}^{2}$

3. C

${\mathrm{m\nu }}^{2}$

4. D

$\frac{1}{6}{\mathrm{m\nu }}^{2}$

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### Solution:

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

dx element's velocity and x's distance from rigid support are,

$\mathrm{v}\text{'}=\frac{\mathrm{vx}}{\mathrm{l}}$

Now the mass of this element of thickness $\left(\mathrm{dx}\right)=\mathrm{\omega dx}=\mathrm{dm}$

Kinetic energy of particle of thickness dx = $\frac{1}{2}dm{v}^{2}$

$\mathrm{KE}={\int }_{0}^{\mathrm{L}}\frac{1}{2}\mathrm{\omega dx}{\left(\frac{\mathrm{vx}}{\mathrm{L}}\right)}^{2}$

$\mathrm{KE}=\frac{1}{2}\frac{\mathrm{m}}{\mathrm{L}}\frac{{\mathrm{v}}^{2}}{{\mathrm{L}}^{2}}{\int }_{0}^{\mathrm{L}}{\mathrm{x}}^{2}\mathrm{dx}$

$\mathrm{KE}=\frac{{\mathrm{mv}}^{2}}{2{\mathrm{L}}^{3}}{{\left[\frac{{\mathrm{x}}^{3}}{3}\right]}^{\mathrm{L}}}_{0}$

$\mathrm{KE}=\frac{1}{6}{\mathrm{mv}}^{2}$

Hence the correct answer is $\frac{1}{6}m{v}^{2}.$

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