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Pulley

Two blocks of...

Question

Two blocks of masses $m_{1}$ and $m_{2}$ are connected with a light spring of force constant k and the whole system is kept on a frictionless horizontal surface.The masses are applied forces $F_{1}$ and $F_{2}$ as shown in $F i g . 6.379$ . At any time, the blocks have same acceleration $a_{0}$ but in opposite directions. Now answer the following:
If $F_{2}$ is removed at this moment, then just after this, the acceleration of $m_{2}$ is
If $F_{2}$ is removed at this moment, then just after this, the acceleration of $m_{2}$ is

\frac{F_{1}}{m_{2}} - a_{0}

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a_{0} + \frac{F_{1}}{m_{2}}

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\frac{F_{2}}{m_{2}} - a_{0}

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a_{0} + \frac{F_{2}}{m_{2}}

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Solution

The correct option is C $\frac{F_{2}}{m_{2}} - a_{0}$
Let, x be the compression in the spring
$F_{1} - k x = m_{1} a_{0}$ ( 1 )
$F_{2} - k x = m_{2} a_{0}$ ( 2 )

when $F_{2}$ suddenly removed spring force acts on $m_{2}$
So, $k x = m_{2} a$
From eq ( 2 )
$k x = F_{2} - m_{2} a_{0}$
$m_{2} a = F_{2} - m_{2} a_{0}$

$a = \frac{F_{2}}{m_{2}} - a_{0}$
$o p t i o n C i s c o r r e c t$

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The correct option is C F2m2−a0Let, x be the compression in the springF1−kx=m1a0 ( 1 )F2−kx=m2a0 ( 2 )when F2 suddenly removed spring force acts on m2So, kx=m2aFrom eq ( 2 ) kx=F2−m2a0 m2a=F2−m2a0 a=F2m2−a0 optionCiscorrect