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))`

Explanation

(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`