The elevator shown in figure is descending with an acceleration of 2ms−2. The mass of the block A=0.5kg. The force exerted by the block A on the block B is (g=10ms−2)
A
2N
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B
4N
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C
6N
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D
8N
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Solution
The correct option is B4N Drawing FBD of ′A′
Here psuedo force acts upwards as lift is moving downwards with acceleration. a=2ms−2 ⇒N+ma=mg N=0.5(10−2)=4N