Calculating Work & Friction for a Dog Pulling a Sled

[jongro]

Homework Statement

This is a question about work:
There is a dog pulling a sled across a patch of snow *with a force of 20 Newtons*, the dog's mouth is 15 degrees above horizontal and the sled weighs 5Kg. The snow has a coefficient of friction of 0.2, how much work has the dog done once it has covered 20 meters in distance (ignore the weight of the dog)?

Homework Equations

W = F_d cos(angle) d

Where:
W is work done
F_d is force applied (20 N)
angle is angle from horizontal (15 degrees)
d is distance (20 meters)

F_f = U * N

Where:
F_f is friction force
U is coefficient of friction (0.2)
N is normal force

The Attempt at a Solution

What I did is:
- Calculated the Normal Force (N)
- Using N, I calculated the force of friction
- Then I calculated the work done by friction (assuming that the angle of the force is 180 degrees)
- Then I calculated the work done by the dog pulling the sled (with 15 degree angle above horizontal)
- Then I added the two (194 + 386)
Result: 580 Joules

Am I doing it right? I feel like I'm in the wrong direction.
Thanks.

 

[javierR]

Well, there seems to be something missing in your setup...just looking at the sliding motion over 20 meters, we'd have to know how much force the dog is applying, but we don't know this unless the sled is said to be moving at constant speed. Then the forces on the sled must balance in the horizontal direction. That means the frictional force F_{f} is equal and opposite to the horizontal part of the force that the dog applies, which I'll call F_{x}. That is, F_{f}+F_{x}=0. But this horizontal part is the only part that matters in the work-by-dog equation: W=F_{x}d since that's the direction of the sled's actual movement (can you see why this is same as using the cos(theta) equation? But in the way I wrote it, do we even need the 15 degree angle?). So now you should be able to piece all this together to get just the work done *by the dog*.

 

[jongro]

Sorry, I forgot the force:
It's 20 Newtons from the dog's mouth (15 degrees form horizontal).