![m = 150 kg u-56 kmlhe-> 15.555 ml r -o t = 12 see v=utat o=is.555+ a 12) sa=(1.296] m/s² Errate - tma = - 150 x 1.296 5-194.4](http://img.homeworklib.com/questions/b9f0dd60-32d6-11ea-a332-1d9b9e9c6bf7.png?x-oss-process=image/resize,w_560)

Cumulative Problem 1 A 150.0-kg crate rests in the bed of a truck that slows from...
A 250.0-kg crate rests on the bed of a dump truck. If the bed of the truck must be raised to an angle of 34.0° for the crate to just begin sliding, what is the coefficient of static friction between the crate and the bed?
A 40.0 kg crate rests on the flat bed of a truck. What is the minimum coefficient of friction between the bed of the truck and the crate for the truck to accelerate at 3.43 m/s2 without the crate sliding along the bed?
Please offer a step by step answer of how you came to know each.
It needs to be very detailed so I can come to understand the
concepts better and reasoning as to why each step was done.
Thanks!
-10 points My Notes A 150-kg crate rests in the bed of a truck that slows from 50 km/h to a stop in 12 s. The coefficient of static friction between the crate and the truck bed is 0.655 (a) Will...