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Question

The value of frictional force and acceleration of block of mass 10 kg in the figure are
(Take g=10 m/s2)


A
10 N,1 m/s2
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B
20 N,2 m/s2
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C
20 N,0 m/s2
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D
10 N,0 m/s2
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Solution

The correct option is C 20 N,0 m/s2
Writing force equation using the FBD of the blocks,
For 4 kg block, T=4g
For 10 kg block in vertical direction,
N+Tsin60=10g
N=10gTsin60
N=10g2g3

fmax=μsN
=0.4[10g23g]
=4083 N

Horizontal force Tcos60=20 N<fmax
Hence frictional force = 20 N and system continues to be at rest.

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