The frictional force acting on $1 \,kg$ block is .................. $N$
A$0.1$
B$2$
C$0.5$
D$5$
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A$0.1$
a (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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