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A block of mass M is kept in elevator (lift) which starts moving upward with constant acceleration b as shown in figure. Initially elevator at rest. The block is observed by two observers A and B for a time interval `t=0` to `t=T`. Observer B is at rest with respect to elevator and observer A is standing on the ground. Q. The observer A finds that the work done by gravity on the block isA. `(1)/(2)Mg^(2)T^(2)`B. `-(1)/(2)Mg^(2)T^(2)`C. `(1)/(2)MgbT^(2)`D. `-(1)/(2)MgbT^(2)` |
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Answer» Correct Answer - D displacement of block in the time interval `T` is `(1)/(2)bT^(2)` `W_("gravity")=(-mg)(+(1)/(2)bT^(2))=-(1)/(2)mgbT^(2)` |
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