In fig m1 = 10 kg and m2 = 5 kg coefficient of static friction (us = 0.2), coefficient of kinetic friction (uk = 0.10) Then find acceleration of block and tension in string?

In fig m1 = 10 kg and m2 = 5 kg coefficient of static friction (us = 0

1.	In fig m1 = 10 kg and m2 = 5 kg coefficient of static friction (us = 0.2), coefficient of kinetic friction (uk = 0.10) Then find acceleration of block and tension in string?
Answer» We know that MAXIMUM FRICTION is given by F=μ×N.According to the given data, we have,N=5×g=50maximum friction force=μN=.5×50=25 NewtonGiven 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= Mf =20/10a=2 m/s 2 Now applying newton's second law on 5 kg block will give,F−20=maF=30 newton

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In fig m1 = 10 kg and m2 = 5 kg coefficient of static friction (us = 0.2), coefficient of kinetic friction (uk = 0.10) Then find acceleration of block and tension in string?​

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In fig m1 = 10 kg and m2 = 5 kg coefficient of static friction (us = 0.2), coefficient of kinetic friction (uk = 0.10) Then find acceleration of block and tension in string?