Two blocks A and B of masses 6kg and 3kg rest on a smooth horizontal surface as shown in the figure. If coefficient of friction between A and B is 0.4, find the maximum horizontal force which can make them move without separation. (Take g=10m/s2)
A
72N
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B
40N
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C
36N
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D
20N
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Solution
The correct option is C36N
Maximum frictional force on block B is f=uN=umBg =0.4×3×10=12N
Hence maximum acceleration of the block B is a=fmB=123=4m/s2 If both the blocks move without separation then, acceleration of both the blocks should be same. Therefore, Fmax=(mA+mB)a Fmax=(6+3)×4 Fmax=36N