Two trolleys 1 and 2 are moving with accelerations $a_{1}$ and $a_{2}$ , respectively, in the same direction. A block of mass m on trolley 1 is in equilibrium from the frame of observer stationary with respect to trolley $2$ . The magnitude of friction force on block due to trolley is (assume that no horizontal force other than friction force is acting on block).

m (a_{1} - a_{2})

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m a_{2}

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m a_{1}

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Data insufficient

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Solution

The correct option is B $m a_{2}$
R.E.F image
$∙$ Let the acceleration of block of mass with respect to
the observer on trolley 2 be denoted as ${\to a}_{b o}$
$∙$ It is given that the block of mass m is in equilibrium with respect to the observer on trolley 2.
$\Rightarrow {\to a}_{b o} = 0$
$\Rightarrow {\to a}_{b} - {\to a}_{o} = 0$ (where ${\to a}_{b}$ denotes the acceleration of block with respect to ground and ${\to a}_{o}$ denotes the acceleration of observer
with respect to ground )
$\Rightarrow_{b} =_{b}$
as $_{o} =_{2} \Rightarrow_{b} =_{2}$
$\Rightarrow$ acceleration of block with respect to ground is
$_{2}$
The friction force on block due to trolley is responsible
for the acceleration $_{2}$
Free body diagram of block
$∴ f = m a_{2}$
$∴$ The magnitude of friction force on block due to
trolley is $f = m a_{2}$

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