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Question

The elevator shown in figure is descending, with an acceleration of 2 m/s2. The mass of the block A is 0.5 kg. The force exerted by the block A on the block B is -
1220627_7defcfdf4f5c4cb68b1c7a2f4b57d6a4.PNG
  1. 2 N
  2. 4 N
  3. 6 N
  4. 8 N

A
4 N
B
8 N
C
6 N
D
2 N
Solution
Verified by Toppr

When the elevator in descending apseudo force acts on it in the upword direction
from free body diagram
m=0.5
g=10
a=2
mgN=ma
N=m(ga)
=0.5(102)=4 N
so the force exerted by the block A m the block B in 4N

1349930_1220627_ans_399b34c36a4541aeacca18c124ab3e6f.png

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