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Question

# The two blocks shown in the figure are not attached. The minimum magnitude of the horizontal force F required to keep the smaller block from slipping down the larger block is - [Take g=10 m/s2]

A
472 N
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B
672 N
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C
372 N
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D
572 N
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Solution

## The correct option is D 572 NLet the combined acceleration of the two blocks system be a. FBD of blocks, From second law of motion,(∑F=ma), for block M N=88a ........(1) [Along horizontal] Similarly, for block m, F−N=16a [Along horizontal] ⇒F−88a=16a [from (1)] ⇒F=104a .........(2) Also, for just slipping condition f=160 [Along vertical] ⇒μN=160 ⇒0.33×88a=160 [from (1)] ⇒a=5.5 m/s2 Putting the value of a in eq.(2) F=104×5.5 ⇒F=572 N is the minimum force required. Why this question ? Concept involved - (1) Newton's law of motions. Tip - If there are two blocks or more which are static with respect to each other, then consider them as a system. Such questions are generally asked in JEE main.

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