The frictional force acting on $1 \,kg$ block is .................. $N$
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(a)

If both move together $a=\frac{10}{101} \simeq 0.1 \,m / s ^2$

Now, $F_{\text {net }}=1(0.1)=0.1 \,N$

$f_L=(0.5)(1)(g)=5 \,N$

So, $f=0.1 \,N$

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