Question

Two blocks A and B of masses 2kg and 3kg are placed on a horizontal surface. The coefficient of friction between A and horizontal surface is 0.5 while block B is smooth. A force F=2^1/2 t N is applied on block A as shown.​

Answer 1

Maximum force is less than 4kg wt.

Explanation:

block B is over block A, so the normal force between the blocks will be the weight of block B. =30N.

so the maximum friction force that can be generated is f=μ.N=0.5×30=15N

so the maximum acceleration of block B is  

mf​ = 315​ =5m/s 2

so the minimum acceleration of both the blocks can be 5m/s2

if both blocks don't want to move together.

so the minimum force required for relative motion is m.a=8×5=40N

means force should be less than 40N, converting to (in kg wt.)

when we divide by 10) we get a maximum force is less than 4kg wt.

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