Question

A force f=kt is applied on 2 kg block at t=0. The displacement of 8 kg block till the time when 2 kg block starts slipping on 8 kg block will be

Answer 1

Answer : The force acting on block =2T

So the max value of friction force acting on block is 0.2*2*10=4N

The max acc of lower block is 4/8=1/2 m/s^2

So we have to find the value of a till t=2 secs as from that point slipping begins

Here the acc is not constant

So dv/dt =t/4

Integrating we get v =t^2/8

S=t^3/24

Substitute value

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