A block of mass 2 kg is placed on the floor . The coefficient of static friction is `0.4` . If a force of `2.8` N is applied on the block parallel to floor , the force of friction between the block and floor is : (Taking g = `10 m//s^(2)`)

Options

A. 2.8 N
B. 8 N
C. 2N
D. zero

Answer

A

Explanation

(a) `f_("max")=mu mg=0.4xx2xx10=8N`
Since the applied force is less then `f_("max")`, force of friction will be equal to the applied force or 2.8 N.