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A body of mass m hangs by an inextensibl...

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 released, calculate the maximum elongation of the spring.

Text Solution

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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 `2x`. Since the spring is elongated by x,
`DeltaPE_(spri ng)=1/2kx^2`
As the body falls through a distance `2x`,
`(DeltaPE)_(gr)=-mg(2x)`
Since the body is initially at rest and comes to rest just at the instant of maximum elongation of the spring, `DeltaPE=0`
Applying the principle of conservation of energy, we get
`(DeltaPE)+(DeltaKE)=0`
`impliesDeltaPE_(Spri ng)+DeltaPE_(gr)+DeltaKE=0`
`1/2kx^2-mg(2x)+0=0impliesx=(4mg)/(k)`
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