# Work question

1. Apr 5, 2005

### tony873004

Two horses on either side of a canal pull a canal boat of mass 1.0*10^4 kg at a constant speed for a distance of 10.0 km. (Figure 7.32). One horse exerts a force of 3.0*10^2 N at an angle of 20º to the canal, and the other exerts 5.0 * 102 N. Find the work done by each horse and the work done by friction between the boat and the water.

I believe I would set it up like this:
Call the direction parallel to the canal the x axis
The work of a horse whose angle is given will be the x-component of the force * distance in meters
$$W_{horse1}=cos(20)*3*10^2N*10000m$$

The work done by the horse whose angle is not given will be the same as the first horse. Again, it will be the x-component of the force, which will be equal to Horse 1's x-component, and although the question doesn't ask for it, I could now compute Horse 2's angle.

But the work done by the friction between the boat and the water? I don't know how to solve that? What if the question said they were on ice instead of water?

The best I can come up with is
$$W_{friction}=Force_{friction} * 10000 meters$$

2. Apr 5, 2005

### Locrian

First, your assumption that the forces in the x-direction does not seem sound to me. On the contrary, I would expect the forces in the y-direction to be the same, but with opposite sign, and then the force in the x-direction of the second horse to be determined after that. I am missing the figure, and also I think there is a mistype, but am not sure (is the second horse actually exerting 5.0 * 10^2N?), so I may not have all necessary information.

Secondly, as to your question about the work that friction does, I hope you don't mind me asking this question: If an object is at a constant velocity for a period of time, what is the net work that has been done on it? Or better yet, if this object is at a constant velocity, what is the net force on it, and what does that mean for the work done?

Hope this helps.

3. Apr 5, 2005

### Gale

To find the force of friction use

$$\Sigma F= ma$$

if the velocity is constant, then acceleration=0.

 i'm too slow...

4. Apr 5, 2005

### tony873004

Thanks, that does help. You're right about the typo 5.0*102 = 5.0*10^2.

Thinking about it, it does make more sense that the y-components should be equal or the boat will hit the side of the canal.

So after setting the y's equal, I have to compute both horses' x-components of force.

You're right, the constant velocity is the key. Water friction must equal the horses pulls but in the opposite direction or there'd be acceleration.

Thanks Locrain!!

5. Apr 5, 2005