InterviewSolution
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InterviewSolution
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A body of mass 0.40 kg moving initially with a constant speed of `10 m//s` to the north is subjected to a constant force of 8.0 N directed towards the south for 30 s Take the instant the force is applied to be t = 0, and the position of the particle at that time to be x = 0, predict its position at `t = -5 s , 25 s, 100 s` ? |
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Answer» Here, m = 0.40 kg , u = 10 `ms^(-1)` , F = -8 N (retarding force) As a = F / m = 8 / 0.4 = - 20 `ms^(-2)` Also , S = `ut + 1/2 aT^(2)` (i) Position at t = - 5s S = 10 (-5) + 1/2 `xx 0 xx (-5)^(2)` = - 50 m (ii) Position at t = 25 s `S_1` = `10 xx 25 + 1/2 xx (-20) xx (25)^(2)` = - 6000 m = - 6 km (iii) Position at t = 100 s `S_2` = ` 10 xx 30 + 1/2 xx (-20) xx (30)^(2)` = - 8700 m At t = 30 s , v = u + at v = 10 - 20 `xx` 30 = -590 `ms^(-1)` Now , for motion from 30 s to 100 s `S_3` = - 590 `xx 70 + 1/2(0) xx (70)^(2)` = - 41300 m Total distance = `S_2 + S_3` = - 8700 - 41300 = - 50000 m = - 50 km . |
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