Coefficient of friction between the flat bad of the truck and crate are `mu = 0.8 and mu_(k) = 0.7`. The coefficient of kinetic friction between the truck tires and the road surface is `0.9` .If the truck stops from an initial speed of `15 m//s` with maximum braking (wheels skidding). Determine where on the bed the crate finally comes to rest. (take `g = 10 m//s^(2))`

Explanation

Assuming that mass of truck gt gt mass of crate . Retardation of truck
`a_(1) = (0.9) g = 9 m//s^(2)`
Retardation of truck
`a_(2) = (0.7) g = 7 m//s^(2)`
or relative acceleration of ceate
`a_(r) = 2 m//s^(2)`
Truck will stop after time
`t_(1) = (15)/(9) = 1.67 s`
and crate will strike the wall at
`t_(2) = sqrt((2 s)/(a_(r))) = sqrt((2 xx 32)/(2)) = 1.78 s`
As `t_(2) gt t_(1)`, crate will come to rest after travelling a distance
`s = (1)/(2) a_(r) t_(1)^(2) = (1)/(2) xx 2.0 xx ((15)/(9))^(2)`
`= 2.77m`