The contact force between 2 kg and 3 kg block placed on an inclined plane as shown in the figure will be
A
3 N
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B
6 N
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C
12 N
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D
18 N
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Solution
The correct option is C12 N Considering both the block as a 5 kg system because both will move the same direction.
Using Newton’s second law: N=5gcos37°(i) 5gsin37°−20=ma=5a(ii) ⇒(5×10×35)−20=5a⇒a=2 m/s2
(Down the inclined plane)
Let the contact force between 2kg and 3kg block be N1. F.B.D. of 3kg block
From Fnet=ma ⇒3gsin37°−N1=3×2 ⇒18−N1=6 ⇒N1=12N