Question

8. 2 kg M = 0.4 4 kg u = 0.5 The minimum value of force F to move the lower block is (1) 32 N (2) 34 N (3) 36 N (4) 38 N​

Answer 1

Answer:

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Explanation:

We know that maximum friction is given by F=μ×N.

According to the given data, we have,

N = 5 × g = 50

maximum friction force = μN =0.5×50=25 Newton

Given friction force is 20  Newton which is less than maximum friction force.

Therefore it's static case which means both of them will move together.

Acceleration of 10 kg Block is given by,

a= f / M  =20/10

a=2 m/s^2

 

Now applying newton's second law on 5 kg block will give,

F − 20 = ma

F = 30 newton