Resistive Force Homework: Find Work & Determine Conservatism

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Homework Help Overview

The discussion revolves around a problem involving a bicyclist experiencing resistive forces while riding in opposite directions. The problem requires calculating the work done by these forces and determining whether the resistive force is conservative.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the calculation of work using the formula W=F*D, while questioning the sign of the forces based on their direction relative to displacement.
  • Some participants express confusion regarding the concept of negative work and how it relates to the definition of conservative forces.
  • There is discussion about whether both forces should be treated as negative due to their opposing directions to displacement.
  • Participants question the implications of their calculations on the classification of the resistive force as conservative or not.

Discussion Status

The discussion is ongoing, with participants providing hints and clarifications about the nature of work and the definitions of conservative forces. There is recognition of the complexity involved in determining the signs of forces and displacements, and some participants have arrived at numerical answers while still grappling with the underlying concepts.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can share or the methods they can use. There is an acknowledgment of potential confusion regarding the treatment of forces and displacements in the context of work calculations.

JJones_86
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Homework Statement


A bicyclist rides 7.8 km due east, while the resistive force from the air has a magnitude of 2.4 N and points due west. The rider then turns around and rides 5.3 km due west. The resistive force from the air on the return trip has a magnitude of 4.5 N and point due east.

a) Find the work done by the resistive force during the round trip.
_____ J
b) Based on your answer to part (a), is the resistive force a conservative force?
Yes
No

Homework Equations


W=F*D

The Attempt at a Solution


W1=2.4N*7800M = 18720 J
W2=4.5N*5300M = 23850 J

Total Work = W1+W2 = 42570 J

I know my setup is wrong, can someone give me some hints? Thanks
 
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JJones_86 said:

Homework Statement


A bicyclist rides 7.8 km due east, while the resistive force from the air has a magnitude of 2.4 N and points due west. The rider then turns around and rides 5.3 km due west. The resistive force from the air on the return trip has a magnitude of 4.5 N and point due east.

a) Find the work done by the resistive force during the round trip.
_____ J
b) Based on your answer to part (a), is the resistive force a conservative force?
Yes
No

Homework Equations


W=F*D

The Attempt at a Solution


W1=2.4N*7800M = 18720 J
W2=4.5N*5300M = 23850 J

Total Work = W1+W2 = 42570 J

I know my setup is wrong, can someone give me some hints? Thanks

Why do you think that is wrong? Work is the integral of the scalar product of force and displacement isn't it?

What about part b?
 
For part a, your setup is mostly not right. However, keep in mind that if a force is applied and the displacement is in the opposite direction, then that force is doing negative work. (I.E. if you lift something straight up, gravity is doing negative work)
 
LowlyPion said:
Why do you think that is wrong? Work is the integral of the scalar product of force and displacement isn't it?

What about part b?
Well I know its not right because I'm getting the wrong answer

Mattowander said:
For part a, your setup is mostly not right. However, keep in mind that if a force is applied and the displacement is in the opposite direction, then that force is doing negative work. (I.E. if you lift something straight up, gravity is doing negative work)
Ok, but am I using the right equations?
 
Yes you seem to be using the right equation. Check your sig figs if you're getting the wrong answer.
 
Mattowander said:
Yes you seem to be using the right equation. Check your sig figs if you're getting the wrong answer.
Ok so are you saying that my Forces should be negative since it's going in the opposite direction?
 
It depends on which direction you take to be negative. What I'm saying is that in both cases, the WORK done by the restrictive force must be negative because the direction of the restrictive forces is opposite the direction of displacement.
 
Mattowander said:
It depends on which direction you take to be negative. What I'm saying is that in both cases, the WORK done by the restrictive force must be negative because the direction of the restrictive forces is opposite the direction of displacement.

Is it possible to have a negative joule though?

W1=-2.4N*7800M = -18720 J
W2=-4.5N*5300M = -23850 J

Im sorry, I'm just really confused right now.
 
  • #10
Mattowander said:
Yes it is possible to have a negative Joule.

Reference this thread for more information : https://www.physicsforums.com/showthread.php?t=53554

I hope that helps.


Ok, so my answer would be..

W1=-2.4N*7800M = -18720 J
W2=-4.5N*5300M = -23850 J

Total Work = W1+W2 = -42570 J?
 
  • #11
The number is right. However, not both of the forces are negative.

If you define East to be positive, then the first 2.4N will be negative. However, the second force 4.5N is pointing East and so must also be positive. When the bike rider changes direction, he is first going towards a positive direction and then towards a negative direction. That is why the Work in either case is negative.
 
  • #12
Mattowander said:
The number is right. However, not both of the forces are negative.

If you define East to be positive, then the first 2.4N will be negative. However, the second force 4.5N is pointing East and so must also be positive. When the bike rider changes direction, he is first going towards a positive direction and then towards a negative direction. That is why the Work in either case is negative.

Oh ok, i see

So it would be
W1=2.4N*7800M = 18720 J
W2=-4.5N*5300M = -23850 J

Total Work = W1+W2 = -5130 J

But if I were to make the 2.4N the negative force, and the 4.5 positive, i would get a +5130 J, so would the final answer just be 5130 J, no matter the sign?
 
  • #13
What I'm saying is that the forces are not always negative. The DISPLACEMENTS however can be negative. If you define East to be the positive x direction, then traveling west would mean your displacement is negative.

W1=-2.4N*7800M = -18720 J
W2=4.5N*-5300M = -23850 J
Total Work = W1 + W2 = -42570J

You had the right answer before but you arrived at it using wrong assumptions. I just wanted to make sure you understand that whether or not the displacement or force is negative depends upon the direction it is in.
 
  • #14
Mattowander said:
What I'm saying is that the forces are not always negative. The DISPLACEMENTS however can be negative. If you define East to be the positive x direction, then traveling west would mean your displacement is negative.

W1=-2.4N*7800M = -18720 J
W2=4.5N*-5300M = -23850 J
Total Work = W1 + W2 = -42570J

You had the right answer before but you arrived at it using wrong assumptions. I just wanted to make sure you understand that whether or not the displacement or force is negative depends upon the direction it is in.


Ok, I feel dumb now, lol. So the answer is -42570J, so the answer to part B is No, right? Because conservative would mean the difference would be 0, correct?
 
Last edited:
  • #15
Assuming that sig figs do not count then yes, both those answers seem correct.
 
  • #16
Mattowander said:
Assuming that sig figs do not count then yes, both those answers seem correct.

Ok, thank you very much for your help, I really appreciate it!
 
  • #17
No problem, anytime.
 
  • #18
I have the same problem with different numbers. The distance is 5.0 km and the resistive force is 3.0 N. I solved using the equation W=(Fd)2 and got 30,000 J, but I do not understand how to determine if the resistive force is a conservative force. Help?
 

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