AIIMS2014PhysicsLaws of Motion
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 the floor, the force of friction between the block and the floor is (g=10 ~m ~s ⁻² )
Options
- A2.8 ~N
- B2 ~N
- C8 ~N
- DZero
Correct answer
A. 2.8 ~N
Step-by-step solution
As f_ m s = _s m g=0.4 2 10=8 ~N . The applied force of 2.8 ~N is less than f_ m s (=8 ~N ) and as such the block does not move. When the block is at rest, frictional force = applied force =2.8 ~N