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Standard XII

Physics

V-t to a-t Graph

In the ideal ...

Question

In the ideal pulley-particle system shown, the mass $m_{2}$ is connected with a vertical spring of spring constant $K$ . $(m_{2} > m_{1})$ . If the mass $m_{2}$ is released from rest when the spring is undeformed, find the maximum compression of the spring.

\frac{(m_{2} - m_{1}) g}{K}

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\frac{(m_{2} + m_{1}) g}{K}

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\frac{m_{1} g}{K}

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\frac{2 (m_{2} - m_{1}) g}{K}

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Solution

The correct option is A $\frac{(m_{2} - m_{1}) g}{K}$
at maximum compression acceleration =0
$T = m_{2} g$
$T = m_{1} g + k x$
$x = \frac{(m_{2} - m_{1}) g}{k}$

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In the ideal pulley-particle system shown, the mass m2 is connected with a vertical spring of spring constant K. (m2>m1). If the mass m2 is released from rest when the spring is undeformed, find the maximum compression of the spring.

The correct option is A (m2−m1)gKat maximum compression acceleration =0T=m2gT=m1g+kxx=(m2−m1)gk

In the ideal pulley-particle system shown, the mass $m_{2}$ is connected with a vertical spring of spring constant $K$ . $(m_{2} > m_{1})$ . If the mass $m_{2}$ is released from rest when the spring is undeformed, find the maximum compression of the spring.

The correct option is A $\frac{(m_{2} - m_{1}) g}{K}$
at maximum compression acceleration =0
$T = m_{2} g$
$T = m_{1} g + k x$
$x = \frac{(m_{2} - m_{1}) g}{k}$