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PhysicsNLMFBD & EquilibriumMedium2 minQB

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The system of two weights with masses $m_{1}$ and $m_{2}$ are connected with weightless spring as shown. The system is resting on the support S . The support S is quickly removed. The accelerations of each of the weights right after the support S is removed are.

Options:

Answer:

Solution:

Initially $m_{1} g = kx$

When support is removed, spring force does not change.

For $m_{1} : m_{1} g - kx = m_{1} a_{1} \Rightarrow a_{1} = 0$ For $m_{2} : m_{2} g + kx = m_{2} a_{2} \Rightarrow a_{2} = \frac{( m _{1} + m _{2} ) g}{m _{2}}$ ↓

Stream:JEESubject:PhysicsTopic:NLMSubtopic:FBD & Equilibrium

⏱ 2mℹ️ Source: QB

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