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In the diagram shown, the normal reaction force between the 2 kg and the 1 kg block is (consider the surface to be smooth). Given g=10 m s⁻²

Asked in NEET 2022 Re-exam · Blocks sliding on a fixed incline

Figure: Blocks sliding on a fixed incline
Answer: (2) 25 N

Step-by-step solution

Given: three blocks on a smooth 30° incline, 3 kg lowest, then 2 kg, then 1 kg at the top; F₁=60 N pushes up the slope at the bottom and F₂=18 N pushes down the slope at the top.

Idea: get the acceleration of the whole set first, then isolate the 1 kg block.

Total mass =3+2+1=6 kg, and the weight component down the slope is 6×10×sin 30°=30 N.

Whole system up the slope: F₁-F₂-Mg sin 30°=Ma, so 60-18-30=6a.

a=(12)/6=2 m s⁻² up the slope.

Now the 1 kg block alone. Up the slope it feels the normal push N from the 2 kg block; down the slope it feels F₂ and its own mg sin 30°=5 N.

N-18-5=1×2.

N=25 N.

Why the other options are wrong

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