JEE Main 2014 · 12 Apr · Shift 1 · Physics Q5

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PhysicsSingle correct+4 / -1Question 5 of 90

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 :

\dfrac12mv^2

mv^2

\dfrac13mv^2

\dfrac16mv^2

Answer and worked solution

Official answer: D

An element at distance $x$ from the fixed end moves with speed $\frac xlv$ . With mass $\frac mldx$ :
$K=\int_0^l\frac12\frac mldx\left(\frac{xv}l\right)^2=\frac{mv^2}{2l^3}\cdot\frac{l^3}3=\frac16mv^2.$

Topics:Work-energy theorem

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JEE Main 2014 · 12 Apr · Shift 1, Physics question 5