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In the arrang...

Question

In the arrangement shown in the figure, all the surfaces are frictionless. The masses of the blocks are $m_{1} = 20 k g$ and $m_{2} = 30 k g$ . If $F = 180 N$ , then the ratio of acceleration of mass $m_{2}$ to that of $m_{1}$ will be

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Solution

$F B D$ of the given diagram

We can see that block $m_{2}$ will remain stationary as there will not be any direct forces were acting, hence it's acceleration will be $0$ .
$m_{1} = 20 k g$
$m_{2} = 30 k g$
For $m_{1}$ :
$T = m_{1} a_{m_{1}}$
$\Rightarrow 20 a_{m_{1}} = 180$
$\Rightarrow a_{m_{1}} = 9 m / s^{2}$
Therefore,
$\frac{a_{m_{2}}}{a_{m_{1}}} = 0$
Final Answer: $0$

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