    Question

# In the arrangement shown in the figure, the wall is smooth and the coefficient of static friction between the blocks is μ=0.1. A horizontal force F=1000 N is applied on the 2 kg block. Identify the wrong statement from the options given below : A
The normal reaction force between the blocks is 1000 N.
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B
The friction force between the blocks is zero.
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C
Both the blocks accelerate downwards with acceleration g.
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D
Friction force acting on 2 kg block is 2g N.
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Solution

## The correct option is D Friction force acting on 2 kg block is 2g N.From the situation shown in figure, it is clear that the blocks cannot accelerate along horizontal direction. FBD of both blocks, From the FBD the equation for horizontal equilibrium of both blocks is, N=F & N′=N Therefore, the normal reaction force between the blocks as well as between 5 kg block and the wall will be same. ∴N′=N=F=1000 N We find that only their respective weight is the external force acting on them in vertical direction. From Newton's second law, a=Fextm a2 kg=2g2=g ; a5 kg=5g5=g Thus, both the blocks will be moving downwards with common acceleration g. Hence, there will be no relative motion between them because both will start from rest with same acceleration. So, there will be no frictional force acting on the blocks because friction always opposes relative motion. Thus, both the blocks accelerate downward with a common acceleration g ms−2 and therefore the relative acceleration between the blocks is zero. Therefore, option (d) is correct. Why this question ?Tip: If two blocks to move together witha common acceleration, the frictional forceacting on the blocks must be less than orequal to the limiting static friction betweenthe blocks i.e., 0≤f≤fl  Suggest Corrections  0      Explore more