First slide

Mechanical energy conservation

Question

A particle of mass m is attached to two vertical springs as shown in figure. Initially the particle is held in position by an external agent such that the springs are in their natural lengths. Now the particle is released. Then maximum elongation and compression produced in the upper and the lower springs respectively are

Moderate

A

$\frac{m g}{2 K}$
B

$\frac{m g}{3 K}$
C

$\frac{3 m g}{2 K}$
D

$\frac{2 m g}{3 K}$

Solution

Elongation in the upper spring = compression produced in the lower spring = x.
change in kinetic energy $(Δ K)$ of the particle is zero.
change in potential energy of the system is given by
$Δ U = Δ U_{e} + Δ U_{g} = (\frac{1}{2} K . x^{2} + \frac{1}{2} .2 K . x^{2}) + (- m g x)$
In a conservative force system
$Δ K + Δ U = 0 \Rightarrow 0 + [\frac{1}{2} .3 K . x^{2} + (- m g x)] = 0 \Rightarrow x = \frac{2 m g}{3 K}$

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