A box weighing 3.0 kN is placed in the bed of a pick-up truck. The pick-up accelerates uniformly from rest to 40 km/h in a distance of 40 m. In that time the box slides 1.0 m back toward the end of the bed. Find the coefficient of friction between the bed and the box.
v^2 = v0^2 + 2a (x-x0)
F = ma
w = mg
Fk = micro k * N
The Attempt at a Solution
v0 = 0 km/h
v = 40 km/h
x-xo = Delta x = 40 m = 0.04 km
weight = 3 kN.
First, find a:
v^2 = v0^2 + 2 a (x-x0)
40^2 = 0 + 2 a * 0.04
=> a = 2*10^4 km/s^2.
My imagine that we have 3 forces in Free Body Diagrams: Friction Forces, Newton's law and weight of the box.
m*g = 3 kN => m = 3000/9.8 = 306.1 grams = 0.306 kg.
N = ma = 0.306 * (2*10^4) = 6120 kN
but the box slides 1.0 m back toward the end of the bed, so
(0.039km * 6120) / 0.04 = 5967 kN
coefficient of friction between the bed and the box is 5967 N / 6120 N = 0.975 = final answer => INCORRECT
Please help! Thanks!