Question

A block of mass 2kg is placed on a truck as shown in figure .The coefficient of kinetic friction between the block and surface is 0.5. The truck starts from rest and moves with acceleration 8m/s².After how much time the block fall off the truck?

Answer 1

Given:- m=2kg, μ=0.5 a0=8m/s^2.

To find:- Time after which the block will fall of.

Solution:-

• a' = μ * g = 0.5*10 = 5m/s^2
• If a' is less than the acceleration of the truck i.e ao the block will fall off and will move backwards w.r.t to the motion of the truck.
• fmax= μ*N = 0.5*2*10= 10N and a = 5m/s^2.
• Acceleration of the block w.r.t truck in the backward direction is given by:-

       a1= ao-a' = 8-5 = 3m/s^2

       s= u*t+1/2(a*t^2)

       6= 0+0.5*(3*t^2)

       t^2 = 4

       t = 2s.

The block will fall off the truck in 2s.