A spring of unstretched length l has a mass m | vedclass

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(D) We can find the effective kinetic energy of the spring by integrating the kinetic energy of all its infinitesimal mass elements.Let $L$ be the len

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$\frac{1}{6} mv^2$

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(D) We can find the effective kinetic energy of the spring by integrating the kinetic energy of all its infinitesimal mass elements.Let $L$ be the length of the spring. Consider an element of length $dy$ at a distance $y$ from the fixed end.The mass of this element is $dm = (m/L) dy$.The velocity $u$ of this element is proportional to its distance $y$ from the fixed end: $u = (y/L)v$.The kinetic energy $dT$ of this element is $dT = \frac{1}{2} (dm) u^2 = \frac{1}{2} (m/L) dy (yv/L)^2$.Integrating from $y = 0$ to $y = L$:$T = \int_{0}^{L} \frac{1}{2} \frac{m}{L} \frac{v^2}{L^2} y^2 dy$$T = \frac{mv^2}{2L^3} \int_{0}^{L} y^2 dy$$T = \frac{mv^2}{2L^3} [y^3/3]_{0}^{L} = \frac{mv^2}{2L^3} (L^3/3) = \frac{1}{6} mv^2$.

$A$ spring of unstretched length $l$ has a mass $m$ with one end fixed to a rigid support. Assuming the 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

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