A crate of mass 45.0 kg is being transported on the flatbed of a pickup truck. T

Question

A crate of mass 45.0 kg is being transported on the flatbed of a pickup truck. The coefficient of static friction between the crate and the truck's flatbed is 0.370, and the coefficient of kinetic friction is 0.210.

(a) The truck accelerates forward on level ground. What is the maximum acceleration the truck can have so that the crate does not slide relative to the truck's flatbed?
3.63 m/s^2 which WebAssign says is correct.

(b) The truck barely exceeds this acceleration and then moves with constant acceleration, with the crate sliding along its bed. What is the acceleration of the crate relative to the ground?

No clue how to figure this out?

Explanation / Answer

Given Mass of crate, m = 45 kg coefficient of static friction,µs =0.360 coefficient of kinetic friction,µk = 0.230 a)The net force on the truck is F = µs m g ma = µs m g a = 0.360 *9.8 m/s^2 a =3.5 m/s^2 Thus, the acceleration of the truck is 3.5 m/s^2 b) The net force on the crate is F = µk m g m a = µk m g a = 0.230 *9.8 m/s^2 a = 2.25 m/s^2 Thus the acceleration of the crate relative to ground is 2.25 m/s^2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.