Calculating Work Done by Horse in 9.9 minutes

[Zhalfirin88]

Homework Statement

A horse pulls a cart with a force of 175.0 N at and angle of 31.0^o 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?

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 a_x 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?

 

[mgb_phys]

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

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

 

[Zhalfirin88]

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

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

No idea.

 

[mgb_phys]

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

 

[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?

 

[mgb_phys]

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

 

[Zhalfirin88]

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

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

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?

Then why is my answer wrong?

 

[mgb_phys]

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

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

 

[Zhalfirin88]

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

It becomes closer to 1.

 

[mgb_phys]

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

 

[Zhalfirin88]

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

 

[tiny-tim]

Hi Zhalfirin88!

A horse pulls a cart with a force of 175.0 N at and angle of 31.0^o 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?
…
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...

Looks ok to me.

(except you have ridiculously many sig figs! )​

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 a_x then use x = vt + .5at^2?

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.

2) Break up the force into components, but I didn't think you had to with the dot product?

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.

 

[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.

 

[tiny-tim]

Got it!

You used cos 31 radians instead of cos 31º

Everything perfect apart from that! ​

 

[Zhalfirin88]

I'm never using google calculator instead of my own again... :P Good catch, never would've though of that.