Question

A block of mass 2 kg is placed on the floor. The coefficient of static friction is ${\mu }_s$=0.4. If a force of 2.8 N is applied on the block parallel to the floor, the force of friction between the block and floor (taking $g=10m/{sec}^2$)

• A. 2.8N
• B. 8N
• C. 2N
• D. Zero

Answer 1

The correct option is A 2.8N

Friction is a self adjusting force.

the max. value of it here is $\mu mg$ = 0.4x20= 8 N

But since the applied force is less than 8N. So, the friction adjusts its value and new value becomes 2.8N so that block remains in equllibrium.