# Find the coefficient of friction

## Homework Statement

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.

## Homework Equations

v^2 = v0^2 + 2a (x-x0)

F = ma

w = mg

Fk = micro k * N

## The Attempt at a Solution

Given:
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.

We have:

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

Therefore:

coefficient of friction between the bed and the box is 5967 N / 6120 N = 0.975 = final answer => INCORRECT

Last edited:

## Homework Statement

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.

## Homework Equations

v^2 = v0^2 + 2a (x-x0)

F = ma

w = mg

Fk = micro k * N

## The Attempt at a Solution

Given:
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.

We have:

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

Therefore:

coefficient of friction between the bed and the box is 5967 N / 6120 N = 0.975 = final answer => INCORRECT

Im no expert, i worked it out and this is what i got: You deduced the first part right, so
a = 1.5m/s2, then the weight of the box = 3000N, so mg = 3000N, so the force of friction = uk(3000N). If you use Newtons third law the foward force = the reaction force, so it is constant motion. So, the force of friction = Fnet, uk(3000N)=ma, a = the above acceleration and m is obtained from the weight equation Fg=mg, m = 306Kg, plug it into above equation and u shud get uk = 0.15.

Hope this helps:)

Im no expert, i worked it out and this is what i got: You deduced the first part right, so
a = 1.5m/s2, then the weight of the box = 3000N, so mg = 3000N, so the force of friction = uk(3000N). If you use Newtons third law the foward force = the reaction force, so it is constant motion. So, the force of friction = Fnet, uk(3000N)=ma, a = the above acceleration and m is obtained from the weight equation Fg=mg, m = 306Kg, plug it into above equation and u shud get uk = 0.15.

Hope this helps:)

How about "In that time the box slides 1.0 m back toward the end of the bed" ???

How about "In that time the box slides 1.0 m back toward the end of the bed" ???
Its neglected since the same acceleration is needed to move the box 1.0m.

Its neglected since the same acceleration is needed to move the box 1.0m.

You mean the final answer is 0.15 ???

You mean the final answer is 0.15 ???

Yup, looks right:)