another friction q?

Question

A greasy flatbed truck is carrying a crate weighing 2.0 kN. The truck accelerates uniformly from rest to a speed of 40 km/h in a distance of 40.0 m. In that time, the crate slides 1.0 m back toward the end of the truck. Compute the coefficient of friction between bed and box.

Answer 1

use vf^2 - v0^2 = 2ax to get the truck's acceleration

a = vf^2 / 2x

the time that takes is vf/a = 2x / vf

In the reference frame of the truck bed, the crate feels a pseudoforce of ma

The net force felt by the crate is the pseudoforce minus friction:
ma - mu mg

So the acceleration on the crate is a - mu g
= vf^2 / 2x - mu g

The distance covered by an object constantly accelerating from rest in a given time:

D = 1/2 (crate accel)t^2

= 1/2 (vf^2 / 2x - mu g) (2x / vf)^2

Solve for mu

mu = (vf^2 / x - D * (vf / x)^2) / 2g

= (1 - D/x) vf^2 / 2gx

Answer 2

The crate moves 39 m while the truck moves 40 m, so the coefficient of friction is 39/40 = 0.975.