Laws of Motion

JEE Main level, single correct, medium

153 of 210

Question 153 of 210

A system of two blocks of masses m=2kgm=2kg and M=8kgM=8kg is placed on a smooth table as shown in figure. The coefficient of static friction between two blocks is 0.50.5. The maximum horizontal force FF that can be applied to the block of mass MM so that the blocks move together will be (g=9.8ms−2)\left(g=9.8m{s}^{-2}\right)

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Answer: (C) 49N49N

Maximum acceleration of the upper block will be,
amax=μg=0.5×9.8=4.9ms−2{a}_{max}=\mu g=0.5\times 9.8=4.9m{s}^{-2}
Therefore, maximum force that can be applied on the lower block is,
Fmax⁡=(2+8)×4.9=49N{F}_{\max }=\left(2+8\right)\times 4.9=49N

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