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A particle moves in a straight line with retardation proportional to its displacement. Its loss in kinetic energy for any displacement x is proportional toA. `x^(2)`B. `e^(x)`C. xD. `log_(e)x` |
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Answer» Correct Answer - 1 As given in question, retardation (negative acceleration)` a prop x` `rArr " " a=-kx` Where K is a proportionality constant `rArr " " vdv=kxdx` Integrating, we get `int_(v_(i))^(v_(f)) vdv=-int_(0)^(x) kx dx` Where `v_(i)` and `v_(f)` respectively are intial and final velocities of particle `((v^(2))/2)_(v_(i))^((v_(f))=-k((x^(2))/2)_(0)^(2)` `rArr " " (v_(f)^(2))/2-(v_(i)^(2))/2=-k (x^(2))/2` `rArr " " 1/2 mv_(f)^(2) -1/2 mv_(i)^(2) =(-1)/2 mkx^(2)` `rArr " " K.E._("final")-K.E._("initial")=1/2 mkx^(2)` Hence, loss in kinetic energy `prop x^(2)` |
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