Question

A block of mass 2 kg placed on rough horizontal surface having coefficient of friction between the
contacts p s = u k = 0.3. If a force F= (10 -3 t) N is applied on the block as shown in the figure. Then
frictional force acting on the block by the surface at t= 2 s will be (g = 10 m/s2)
A​

Answer 1

Answer:

Since applied force (6N) > max frictional force (1N):

(I'm assuming static friction coefficient = kinetic friction coefficient),

6 - 1 = 2a

a = 5/2 m/s^2.