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# Sir pls solve the following.

Asked by rsudipto 22nd December 2018, 8:06 PM
Figure shows the free body diagram of both blocks.
Limiting friction force acting on 1 kg block along the direction of motion is μR1 = 0.1×2×10 = 2 N.
where μ is the friction coefficient between blocks and R1 is the reaction force acting between the contacts

Acceleration of 1 kg block = 2/1 = 2 m/s² .

If 2 kg block also moving with same acceleration, then we have as per Newton's law =  T - μR1 = m×a = 2×2
where T is the tension force that is pulling the 2 kg block

T = 4 + 2 = 6 N

Hence mass hanging in the pulley = 6/g = 6/10 = 0.6 kg
Answered by Expert 23rd December 2018, 6:51 AM
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