Find work done pulling sled with rope

[ObviousManiac]

Homework Statement

A loaded sled is pulled by means of rope that makes an angle of 45 with the horizontal. If the mass of the sled is 60 kg and the coefficient of friction is .0200, how much work is done in pulling the sled at constant velocity along a level road for a distance of 1.00 km?

Homework Equations

Problem solved here:

http://www.flickr.com/photos/landolukes/5831482036/lightbox/"

The Attempt at a Solution

My final answer is 16,305 J. Yet, My teacher got 11,541 J. My tutor says I did it right, so I thought I'd bring it here for some other opinions.

 

[cepheid]

It's definitely 11,541 J. Don't forget the contribution to the normal force from the vertical component of the applied (pulling) force.

EDIT: In your work, it looks like you forgot that only the component of F_applied that is parallel to the displacement of the sled does any work (i.e. only the x-component does work). So you are missing a factor of cos(45°) from the final answer.

 

[nayfie]

This may be a dumb question, but wouldn't it be 11,772 J?

 

[cepheid]

This may be a dumb question, but wouldn't it be 11,772 N?

Without seeing any work, I can't really comment further. How did you arrive at that answer? I'm more than happy to show what I did, if you like.

 

[nayfie]

W = F \bullet s

There is no displacement in the vertical axis, therefore W_{y} = 0

F = F_{friction} = \mu F_{N} = (0.02)(60*9.81) \approx 11.772

W_{x} = F \bullet s = |F||s|cos\theta = (11.772)(1000)cos0 \approx 11,772 J

 

[cepheid]

Your error comes in the third line where you assume that F_N = mg. The normal force is slightly less than the weight, because the person is pulling upward on the box with a vertical force equal to Fsinθ. Hence, the magnitude of the normal force is mg - Fsinθ.

 

[nayfie]

Oh yes! How obvious...

Thank you :)