Question : A block of mass 2 kg is placed on a plane surface. The coefficient of static friction is 0.4. When a 2.8 N force is applied on the block parallel to the surface, the force of friction between the block and the surface is
(a) 2.8 N
(b) 7.84 N
(c) 0
(d) 9 N
Doubt by Ishmeet
Solution :
Mass of the body (m) = 2 kg
Coefficient of friction (µ) = 0.4
Applied Force (F) = 2.8 N
fs=µR
fs=µmg
fs=(0.4)(2)(9.8)
fs=µmg
fs=(0.4)(2)(9.8)
fs=7.84 N
but F=2.8 N
F<fs
but F=2.8 N
F<fs
Since the applied force is less than maximum friction force. So the block will not move because it will provide a friction force equal and opposite to the applied force i.e. 2.8 N.
Hence, (a) 2.8 N, would be the correct option.
Similar Question :
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 (Take g=10 m/s²)
(a) 2.8 N
(b) 8 N
(c) 2 N
(d) Zero