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Pulley Problem

Two blocks of...

Question

Two blocks of masses $M_{1}$ and $M_{2}$ are connected to each other through a light spring as shown in figure. If we push mass $M_{1}$ with force F and cause acceleration $a_{1}$ in right direction for mass $M_{1}$ , what will be the magnitude of acceleration of $M_{2}$ ?

F / M_{2}

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F (M_{1} + M_{2})

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a_{1}

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(F - M_{1} a_{1}) / M_{2}

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Solution

The correct option is D $(F - M_{1} a_{1}) / M_{2}$
Let x be the compression in the spring.The equation of motion of the blocks are:
$F - k x = M_{1} a_{1}$ .........................(1)
$k x = M_{2} a_{2}$ .........................................(2)
Adding these equations we get $F = M_{1} a_{1} + M_{2} a_{2}$
$∴$ $a_{2} = \frac{F - M_{1} a_{1}}{M_{2}}$
hence,option D is correct.

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Two blocks of masses M1 and M2 are connected to each other through a light spring as shown in figure. If we push mass M1 with force F and cause acceleration a1 in right direction for mass M1, what will be the magnitude of acceleration of M2 ?

The correct option is D (F−M1a1)/M2Let x be the compression in the spring.The equation of motion of the blocks are:F−kx=M1a1.........................(1)kx=M2a2.........................................(2)Adding these equations we get F=M1a1+M2a2 ∴ a2=F−M1a1M2hence,option D is correct.

Two blocks of masses $M_{1}$ and $M_{2}$ are connected to each other through a light spring as shown in figure. If we push mass $M_{1}$ with force F and cause acceleration $a_{1}$ in right direction for mass $M_{1}$ , what will be the magnitude of acceleration of $M_{2}$ ?