Calculating Work Done by Friction: Step-by-Step Explanation for Beginners

In summary, the two villains in this scenario exerted equal horizontal forces of 319N to keep the laundry basket moving at a constant velocity of 1.52m/s. The work done by friction was 4880.7 Joules.
  • #1
Chica1975
63
0
I have a physics teacher who is unable to explain anything if his life depended on it hence I am really struggling with physics. I am left out on a limb literally - at my wits end.

I have this question that I have no idea where to start with.

2 villans slid a laundry basket containing a body a distance of 7.76m across a warehouse floor. They both exerted equal horizontal forces of 319N in order to keep the basket moving at a constant velocity of 1.52m/s. How much work was done by friction?

Please explain to me step by step why we get to the answer please don't just give me the equations and the answer I need to understand why we get there. I did math 20 years ago so its a whole new ball game for me.

Thanks
 
Physics news on Phys.org
  • #2
Chica1975 said:
I have a physics teacher who is unable to explain anything if his life depended on it hence I am really struggling with physics. I am left out on a limb literally - at my wits end.

I have this question that I have no idea where to start with.

2 villans slid a laundry basket containing a body a distance of 7.76m across a warehouse floor. They both exerted equal horizontal forces of 319N in order to keep the basket moving at a constant velocity of 1.52m/s. How much work was done by friction?

Please explain to me step by step why we get to the answer please don't just give me the equations and the answer I need to understand why we get there. I did math 20 years ago so its a whole new ball game for me.

Thanks

Welcome to the PF.

What are the relevant equations for calculating work from force and distance? What is the total force in this question? What is the distance?
 
  • #3
Thanks

W = F*d cos theta
total force is 638
distance is 7.65m
 
  • #4
Chica1975 said:
Thanks

W = F*d cos theta
total force is 638
distance is 7.65m

Good. And since the question says that the forces are applied horizontally (parallel with the direction of motion), what are theta and cos(theta)? And your final answer (including units) is... ?
 
  • #5
theta and cos theta (I think) should be 0 because they are parallel?
I did do this and got an answer of 4880.7 Joules - should be negative because friction is going in the opposite direction.

their answer is -4.95 KJ even if I divide by 1000 I still don't get the right answer.
 
  • #6
Chica1975 said:
theta and cos theta (I think) should be 0 because they are parallel?
I did do this and got an answer of 4880.7 Joules - should be negative because friction is going in the opposite direction.

their answer is -4.95 KJ even if I divide by 1000 I still don't get the right answer.

Sorry, I missed a typo in your previous post. The distance in the original problem in your first post is 7.76m, not 7.65m. That accounts for the small difference between your answer and theirs.

BTW, you should have said theta = 0, so cos(0) = 1. That must be what you meant, since you would have gotten the right answer without the distance typo.

Good job.
 
  • #7
thanks a mil - sometimes you just need guidance to get your confidence and when you have been looking at it for hours without success it gets harder and harder.
 

1. What is friction?

Friction is a force that resists the relative motion of two objects in contact with each other.

2. How is work done by friction calculated?

Work done by friction is calculated by multiplying the force of friction by the distance over which it acts.

3. How does friction affect the amount of work done?

Friction always acts in the direction opposite to the motion of an object, so it decreases the amount of work done.

4. Can friction do positive work?

Yes, friction can do positive work if the displacement of the object is in the same direction as the force of friction.

5. How can the work done by friction be reduced?

The work done by friction can be reduced by using lubricants, increasing the smoothness of surfaces, and reducing the weight of the objects in contact.

Similar threads

  • Introductory Physics Homework Help
2
Replies
58
Views
3K
  • Introductory Physics Homework Help
Replies
7
Views
3K
Replies
29
Views
1K
Replies
7
Views
2K
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
898
  • Introductory Physics Homework Help
Replies
19
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
648
  • Introductory Physics Homework Help
Replies
3
Views
934
  • Introductory Physics Homework Help
Replies
7
Views
1K
Back
Top