Question

The frictional force acting on $1kg$ block is

• A. 0.1 N
• B. 2 N
• C. 0.5 N
• D. 5 N

Answer 1

Answer: (A)

If both move together $a=\frac{10}{101}\simeq 0.1m/s^2$
Now,
$F_{\text{net }}=1(0.1)=0.1N$
$f_L=(0.5)(1)(g)=5N$
So, $f=0.1N$