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A body starts from rest and is acted on by a constant force. The ratio of kinetic energy gained by it in the first five seconds to that gained in the next five seconds isA. `2 : 1`B. `1 : 1`C. `3 : 1`D. `1 : 3` |
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Answer» Correct Answer - D `KE = (1)/(2) mv^(2) = (1)/(2) m(g t)^(2) (because v = g t)` `(K_(1))/(K_(o)) = (t_(1)^(2))/(t_(2)^(2) -t_(1)^(2))` where `t_(1) = 5 sec` and `t_(2) = 10 sec`. |
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