Question

# In the arrangement shown in ﬁgure mA=mB=2 kg. String is massless and pulley is frictionless. Block B is resting on a smooth horizontal surface, while friction coefﬁcient between blocks A and B is μ=0.5. What maximum horizontal force F can be applied so that block A does not slip over the block B? (g=10 m/s2)

A

25N

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B

40N

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C

30N

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D

20N

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Solution

## The correct option is D 20N Net horizontal force on block B is zero. Hence, the given figure (a) can be replaced by figure (b). Block B moves due to friction. Therefore, maximum common acceleration of the two blocks can be amax=fmaxmB=102=5 m/s2∴Fmax=(mA+mB)amax=(2+2)(5)=20N

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