Find the coefficient of friction

• huybinhs
In summary, 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. The coefficient of friction between the bed and the box is calculated to be 0.15.
huybinhs

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:

huybinhs said:

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:)

E=mc^84 said:
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" ?

huybinhs said:
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.

E=mc^84 said:
Its neglected since the same acceleration is needed to move the box 1.0m.

You mean the final answer is 0.15 ?

huybinhs said:
You mean the final answer is 0.15 ?

Yup, looks right:)

1. What is the coefficient of friction?

The coefficient of friction is a dimensionless quantity that represents the amount of resistance or force between the surfaces of two objects in contact. It is used to measure how easily one object slides or moves against another.

2. How is the coefficient of friction calculated?

The coefficient of friction is calculated by dividing the force required to move one object over another by the normal force, or the force that is pressing the two objects together. This is typically represented by the equation μ = F/N, where μ is the coefficient of friction, F is the applied force, and N is the normal force.

3. What factors affect the coefficient of friction?

The coefficient of friction can be affected by various factors, such as the nature of the two surfaces in contact, the roughness or smoothness of the surfaces, the temperature, and the presence of any lubricants or contaminants. It can also vary based on the type of motion, such as static or kinetic friction.

4. What are some common applications of finding the coefficient of friction?

The coefficient of friction is commonly used in engineering and physics to design and analyze the motion of objects, such as in the design of brakes, tires, and other moving parts. It is also used in forensic investigations to determine the cause of accidents or injuries, and in sports to analyze the performance of athletes and equipment.

5. How can the coefficient of friction be reduced?

The coefficient of friction can be reduced by using lubricants, such as oil or grease, between two surfaces to reduce the amount of friction and allow for smoother movement. Additionally, changing the surface materials or making them smoother can also decrease the coefficient of friction. In some cases, adding weight or increasing the surface area of an object can also reduce the coefficient of friction.

• Introductory Physics Homework Help
Replies
11
Views
2K
• Introductory Physics Homework Help
Replies
18
Views
2K
• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
10
Views
1K
• Introductory Physics Homework Help
Replies
48
Views
6K
• Introductory Physics Homework Help
Replies
3
Views
716
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
137