InterviewSolution
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InterviewSolution
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A package rest on a conveyor belt which is initially at rest . The belt is started and moves to the right for `1.3 s` with a constant acceleration of `2 m//s^(2)`. The belt then moves with a constant deceleration `a_(2)` and comes to a stop after a total displacement of `2.2 m` . Knowing that the coefficient of friction between the package and the belt are `mu_(s) = 0.35` and `mu_(k) = 0.25` , determine (a) the deceleration `a_(2)` of the belt , (b) the displacement of the package relative to the belt as the belt comes to a stop. `(g = 9.8 m//s^(2))` |
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Answer» (a) `nu = a_(1) t_(1) = 2.6 m//s` `s_(1) = (1)/(2) a_(1) t_(1)^(2) = (1)/(2) xx 2 xx (1.3)^(2) = 1.69 m` `s_(2) = (2.2 - 1.69) = 0.51 m` and `s_(2) = (nu^(2))/(2 a_(2))` `:. a_(2) = (nu^(2))/(2 s_(1)) = (2.6)^(2)/(2 xx 0.51) = 6.63 m//s^(2)` and `t_(2) = (nu)/(a^(2)) = 0.4 s` (b) Acceleration of package will be `2 m//s^(2)` while retardation will be `mu_(k)g or 2.5 m//s^(2) "not" 6.63 m//s^(2)` for the package. `nu = a_(1) t_(1) = 2.6 m//s implies s_(1) = (1)/(2) a_(1) t_(1) = 1.69m` ` s_(2) = nut_(2) - (1)/(2) d_(2) t_(2)^(2) = 2.6 xx 0.4 - (1)/(2) xx 2.5 xx (0.4)^(2)` `= 0.84 m` `:.` displacement of package w.r.t. belt `= (0.84 - 0.51) m = 0.33 m` |
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