How Do You Calculate Work and Kinetic Friction for Constant Velocity?

  • Thread starter Thread starter potential thinker
  • Start date Start date
  • Tags Tags
    Figure
Click For Summary

Homework Help Overview

The discussion revolves around calculating work and kinetic friction for an object moving at constant velocity on a rough surface. The original poster presents a scenario involving a horizontal force applied to a bag, prompting questions about the work done and the coefficient of kinetic friction.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between work and kinetic energy, with one suggesting that work done is equal to the force multiplied by the distance. Others raise questions about the definitions and implications of kinetic friction in this context.

Discussion Status

The discussion is active, with participants providing hints and clarifications regarding the concepts of work and kinetic friction. Some guidance has been offered, particularly around the definitions involved, but no consensus or resolution has been reached.

Contextual Notes

The original poster's questions indicate a need for foundational understanding of work and friction, particularly in the context of constant velocity, suggesting that assumptions about the system may need to be examined.

potential thinker
I can't figure out these questions...HELP!

1. A horizontal force of 150N is used to push a 40kg bag a distance of 6.00m on a rough horizontal surface. If the crate moves with constant velocity:
a. what is the work done by the force
b. what is the coefficient of kinetic friction

How do i start this?
 
Physics news on Phys.org
Just a repeat answer since the symbols didn't come out correctly ...

The Work Done = Change in Kinetic Energy, which means that:

[tex]F * d = m/2(v_{f}^2 - v_{i}^2)[/tex],where,
[tex]v_{i}[/tex] & [tex]v_{f}[/tex] are the initial and final velocities respectively and [tex]F[/tex] is the force and [tex]d[/tex] is the displacement. In your case the initial velocity is zero, hence the work done is directly equal to the kinetic energy.

From this you can find the velocity of the object.

As for your question the work done is directly the Force * distance that the force has pushed the object to.

Kinetic Friction is basically the friction that is exerted by the surface on a body that is moving. Thus the Kinetic Frictional force = [tex]F = \mu_{k}*mg[/tex], where [tex]\mu_{k}[/tex] is the coefficient of kinetic friction and [tex]F[/tex] is the force applied.This is done as the frictional force always acts in a way to oppose motion. Hence the magnitude of the frictional force = magnitude of the applied force.

Sridhar
 
hey sridhar what is [tex]?<br /> thanks for the help[/tex]
 
tex is basically the latex tag that is used to begin typing latex commands that can be used to generate symbols like I have in my 2nd answer. I screwed up on the tags in the previous answer, so I have corrected them in my second Just click on one of those tags to see how the symbols are got(Those Huge black formulae...jus click on one of them...go ahead)

u can use latex here to type something like this:

[tex]\sin^2\Phi+\cos^2\Phi = 1[/tex]

Just click on the Huge Black Formula above to see the Latex code.

Sridhar
 
You can also do something like this using Latex:

[tex]\int_{0}^\infty f(x)\partial x[/tex]

or something like:

[tex]\sum (n) = n(n+1)/2[/tex]


So as you see the display beomes more organised wit latex. Yet the time taken to type an answer using latex is enormous..


Sridhar
 
srihar_n
A hint for you, there is a edit button at the bottom of each message, if you wrote it, you can edit it! :smile:
 


Originally posted by potential thinker
a. what is the work done by the force
b. what is the coefficient of kinetic friction
For part a, consider the definition of work.
For part b, consider what the force of friction must equal since the crate has a constant velocity.
 

Similar threads

  • · Replies 58 ·
2
Replies
58
Views
6K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 33 ·
2
Replies
33
Views
4K
Replies
18
Views
4K
Replies
2
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
26
Views
5K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 16 ·
Replies
16
Views
6K
  • · Replies 5 ·
Replies
5
Views
3K