The converyor belt is designed to transport packges of verious weights.Each `10 kg` package has a coefficient of kinetic friction `mu_(k) = 0.15`.If the speed of the conveyor belt is `5 m//s` and then it suddenly stops, determine the distance the package will slide before coming to rest `(g = 9.8 m//s^(2))`

Explanation

Retardation `a = mu_(k) g = 0.15 xx 9.8 = 1.47 m//s^(2)`
Distance travelled before slidding stop is,
`s = (nu^(2))/(2 a)`
`= (5)^(2)/(2 xx 1.47) = 8.5 m`