The value of frictional force and acceleration of block of mass 10kg in the figure are (Take g=10m/s2)
A
10N,1m/s2
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B
20N,2m/s2
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C
20N,0m/s2
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D
10N,0m/s2
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Solution
The correct option is C20N,0m/s2 Writing force equation using the FBD of the blocks, For 4kg block, T=4g For 10kg block in vertical direction, N+Tsin60∘=10g ⇒N=10g−Tsin60∘ N=10g−2g√3
fmax=μsN =0.4[10g−2√3g] =40−8√3N
Horizontal force Tcos60∘=20N<fmax Hence frictional force = 20N and system continues to be at rest.