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Question

In the figure shown co-efficient of friction between the block B and the block C is 0.4. there is no friction between the block C and the surface on which it is placed. The system of blocks is released from rest in the situation shown in the diagram. Find the distance moved by the block C when block A descends through a distance 2 m. Given masses of the blocks are mA = 3 kg, mB = 5 kg and mC = 10 kg.


  1. 0 m

  2. 5 m

  3. 4 m

  4. 2 m


Solution

The correct option is D

2 m


Let there is no relative motion between the blocks B and C

Hence

T = (mB+mC)a     .........(1)

And

mAgT=mAa        ..........(2)

from (1) and (2), we get

a=mAgmA+ma+mc=3018=53m/s2

  Net  force  on  the block e C is,F=mca=10×(53)N=16.6N

If maximum value of frictional force acting on block C is fmax then

F(max)=μmBg=0.4×10=20NFfmax

Hence there is no relative motion between the block B and C. Therefore, distance moved by C is 2 m only.

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