# Horse Power

1. Oct 20, 2009

### Zhalfirin88

1. The problem statement, all variables and given/known data
A horse pulls a cart with a force of 175.0 N at and angle of 31.0o with respect to the horizontal and moves along at a speed of 10.4 km/hr. How much work does the horse do in 9.9 min?

3. The attempt at a solution

Distance = $${2.889 \frac{m}{s}} * 594s = 1716.066 m$$

$$175 * 1 716.066 * cos 31^o = 274707.695 J$$

But that doesn't even look right either...

I can think of 2 ways why this is wrong.
1) I didn't use the right formula for distance. Would I have to solve for ax then use x = vt + .5at^2?
2) Break up the force into components, but I didn't think you had to with the dot product?

2. Oct 20, 2009

### mgb_phys

It isn't - which side of the triangle are you looking for?

Hint - what happens to the power as the 31deg -> zero?

3. Oct 20, 2009

### Zhalfirin88

No idea.

4. Oct 20, 2009

### mgb_phys

If the slope gets flatter is that more or less work?
Does cos(angle) get larger or smaller as 'angle' gets less?

5. Oct 20, 2009

### DavyCrocket20

Why would it matter what the components of force are? Isn't it just asking how much work the horse is doing? Wouldn't that just be force*distance? Granted, gravity is doing negative work, but are you sure you have to take that into consideration?

6. Oct 21, 2009

### mgb_phys

Exactly - without friction you only do work in the distance the force moves against gravity

7. Oct 21, 2009

### Zhalfirin88

I don't really get how helps, the lesser the slope the lesser the work.

Then why is my answer wrong?

Last edited: Oct 21, 2009
8. Oct 21, 2009

### mgb_phys

Correct,
now how does Cosine() change as the angle gets smaller ?

9. Oct 21, 2009

### Zhalfirin88

It becomes closer to 1.

10. Oct 21, 2009

### mgb_phys

But as the slope gets smaller you want the amount of work to get closer to zero !

11. Oct 21, 2009

### Zhalfirin88

I don't get what you're trying to tell me!!

12. Oct 22, 2009

### tiny-tim

Hi Zhalfirin88!
Looks ok to me.

(except you have ridiculously many sig figs! )​
No, this sort of question is telling you that the speed is a constant 10.4 …

it doesn't bother to tell you why, but obviously there must be some other forces, such as friction, which are reducing the net force to zero, so the acceleration is zero.

Your x = vt formula is correct.
That's right, you don't.

Work done is just force "dot" distance moved …

the force is at 31º to the (horizontal) distance moved, so cos31º should be correct.

13. Oct 22, 2009

### Zhalfirin88

When I typed it online for an answer I put 2.75 x 10^5 J. I don't know why it's wrong, I've given everything the problem says, there wasn't a picture or anything.

14. Oct 22, 2009

Got it!