The elevator shown in the figure is descending with an acceleration of 2m/s2. The mass of the block A=0.5kg. The force exerted by the block A on the block B is (take g=10m/s2)
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 Let A applies a force R on B, then B also applies an opposite and equal force R on A.
The FBDs of the block are as shown
Considering the frame of reference as elevator, a pseudoforce ma will be acting on A upwards where m is mass of block A, therefore mg−R=ma ⇒R=m(g−a)=0.5[10−2]=4N