Question

A body of mass $m$ hangs by an inextensible string that passes over a smooth mass less pulley that is fitted with a light spring of stiffness $k$ as shown in figure. If the body is released from rest and the spring is undeformed, calculate the maximum elongation of the spring.

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Accepted Answer

Let the spring be elongated by $x .$ That means the pulley falls through a distance $x .$ Consequently, the string of the pulley is slackened to the same extent $x .$ Therefore, the total distance moved by the body is $2 x .$ Since the spring is elongated by $x .$

$Δ P E_{spring} = \frac{1}{2} k x^{2}$

As the body falls through a distance $2 x,$

${(ΔPE)}_{gr} = - m g (2 x)$

Since the body is initially at rest and comes to rest just at the instant of maximum elongation of the spring, $Δ KE = 0$ . Applying the principle of conservation of energy, we get,

$(Δ PE) + (Δ KE) = 0$

$\Rightarrow Δ {PE}_{spring} + Δ {PE}_{gr} gr + Δ KE = 0$

$\frac{1}{2} k x^{2} - m g (2 x) + 0 = 0 \Rightarrow x = \frac{4 m g}{k}$

B M Sharma Solutions for Chapter: Work, Energy and Power, Exercise 57: ILLUSTRATION 8.54

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Work, Energy and Power, Exercise 57: ILLUSTRATION 8.54

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