Resistive Force Homework: Find Work & Determine Conservatism

[JJones_86]

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

 

[LowlyPion]

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?

 

[Mattowander]

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)

 

[JJones_86]

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

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?

 

[Mattowander]

Yes you seem to be using the right equation. Check your sig figs if you're getting the wrong answer.

 

[JJones_86]

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?

 

[Mattowander]

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.

 

[JJones_86]

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.

 

[Mattowander]

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.

 

[JJones_86]

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?

 

[Mattowander]

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.

 

[JJones_86]

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 postive, i would get a +5130 J, so would the final answer just be 5130 J, no matter the sign?

 

[Mattowander]

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.

 

[JJones_86]

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?

 

[Mattowander]

Assuming that sig figs do not count then yes, both those answers seem correct.

 

[JJones_86]

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!

 

[Mattowander]

No problem, anytime.

 

[Cluless]

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?