MCQ
The frictional force acting on $1 \,kg$ block is .................. $N$
- ✓$0.1$
- B$2$
- C$0.5$
- D$5$
✓
Answer
Correct option: A.
$0.1$
a
(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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