InterviewSolution
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InterviewSolution
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A bolck of mass `10 kg` is moving in x-direction with a constant speed of `10 m//s`. it is subjected to a retardeng force `F=-0.1 x J//m`. During its travel from `x =20m` to `x =30 m`. Its final kinetic energy will be .A. `250 J`B. `450 J`C. `275 J`D. `475 J` |
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Answer» When a force acts upon a moving body , then the kinekic energy of the body increase and the increase is equal to the work done. This is work energy therom. Work done `= (1)/(2) mv^(2) - (1)/(2) mu^(2) = K_(f) - K_(i)` Another definition of work is force xx displacement. `:. Fdx = K_(f) - (1)/(2) xx mv_(1)^(2)` where the subscripts `f` and l` stand for final and initial. `F. dx = K_(f) - (1)/(2) xx10 xx (10)^(2)` `implies F. dx = K_(f) - 500` `rArr underset(x=20)overset(x=30)int=K_(f)-500` Using the formula `int x^(n) dx = (x^(n+1))/(n+1)`, we have `- 0.1 [(x^(2))/(2)]_(x=20)^(x=30) = K_(f) - 500` `- 0.1 [(30)^(2)/(2) - ((20)^(2))/(2)] = K_(f) - 500` `implies K_(f) - 500 = - 25` `implies K_(f) - 500 - 25 = 475J` |
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