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Question

What should be the maximum value of M so that the 4 kg block does not slip over the 5 kg block: (Take g=10 m/s2)


A
12 kg
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B
8 kg
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C
10 kg
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D
6 kg
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Solution

The correct option is D 6 kg
Static friction (f) will act at its maximum value, so that both blocks 4 kg and 5 kg can move together corresponding to maximum mass of block M.
Hence, system of blocks 4 kg,5 kg and M will move together with acceleration a, due to string constraint.


Net external force on system of blocks (4 kg,5 kg,M) is Mg downwards.
a=FnetTotal mass=MgM+5+4
=MgM+9 ... (i)

From FBD of 4 kg:


N=mg=40 N
& fmax=4a ... (ii)
fmax=μN=0.4×40=16 N
a=164=4 m/s2

Putting in eq.(i) gives:
4=M×10M+96M=36
M=6 kg

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