Question

A block of mass 2 kg is placed on the floor. The coefficient of ststic friction is 0.4. A force F of 3 N is applied on the block as shown in figure. The force of friction between the block and the floor is (Take $g=10m{s}^{-2}$)

• A. 3 N
• B. 8 N
• C. 4 N
• D. 6 N

Answer 1

Answer: (A)

3 N

Given,

mass of block, $m=2kg$

acceleration due to gravity, $g=10m/s^2$

Coefficient of friction, $\mu =0.4$

External applied force on the block, $F=3N$

Limiting friction value, in this case, can be given as:

$F_{max}=\mu mg$

$F_{max}=0.4\times 2\times 10$

$F_{max}=8N$

But as we know the friction is a self-adjusting force when the applied external force used to be less or equal to the limiting friction value.

And, here $F=3N<F_{max}=8N$

Hence, the friction force experienced by the block, in this case, will be, $f=F=3N$ [Applied force]

So, the correct option is $\left(A\right)$