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 -
2 N
4 N
6 N
8 N
A
4 N
B
8 N
C
6 N
D
2 N
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Solution
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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
mg−N=ma
N=m(g−a)
=0.5(10−2)=4N
so the force exerted by the block Am the block B in 4N
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