The frictional force acting on $1 k g$ block is

0.1 N

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2 N

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0.5 N

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5 N

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Solution

The correct option is D $5 N$
$\begin{matrix} m_{1} = 1 k g m_{2} = 100 k g N_{1} = M_{1} g \Rightarrow N_{1} = 1 \times 98 = 9.8 N \end{matrix}$
Both blocks move together with are 'a'
$\begin{matrix} F = m_{1} a f = m_{1} a (f u i t i o n) a l s o, f = M N ∴, M N = 0.5 \times 9.8 = 4.9 \approx 5 N \end{matrix}$

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The frictional force acting on 1 kg block is

The correct option is D 5 Nm1=1kgm2=100kgN1=M1g⇒N1=1×98=9.8NBoth blocks move together with are 'a' F=m1af=m1a(fuition)also,f=MN∴,MN=0.5×9.8=4.9≈5N

The frictional force acting on $1 k g$ block is