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Horse Power

  1. Oct 20, 2009 #1
    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 = [tex]{2.889 \frac{m}{s}} * 594s = 1716.066 m[/tex]

    [tex]175 * 1 716.066 * cos 31^o = 274707.695 J[/tex]

    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. jcsd
  3. Oct 20, 2009 #2

    mgb_phys

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

    Hint - what happens to the power as the 31deg -> zero?
     
  4. Oct 20, 2009 #3
    No idea.
     
  5. Oct 20, 2009 #4

    mgb_phys

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    If the slope gets flatter is that more or less work?
    Does cos(angle) get larger or smaller as 'angle' gets less?
     
  6. Oct 20, 2009 #5
    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?
     
  7. Oct 21, 2009 #6

    mgb_phys

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    Exactly - without friction you only do work in the distance the force moves against gravity
     
  8. Oct 21, 2009 #7
    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
  9. Oct 21, 2009 #8

    mgb_phys

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    Correct,
    now how does Cosine() change as the angle gets smaller ?
     
  10. Oct 21, 2009 #9
    It becomes closer to 1.
     
  11. Oct 21, 2009 #10

    mgb_phys

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    But as the slope gets smaller you want the amount of work to get closer to zero !
     
  12. Oct 21, 2009 #11
    I don't get what you're trying to tell me!!
     
  13. Oct 22, 2009 #12

    tiny-tim

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    Hi Zhalfirin88! :smile:
    Looks ok to me. :confused:

    (except you have ridiculously many sig figs! :rolleyes:)​
    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. :smile:
    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.
     
  14. Oct 22, 2009 #13
    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.
     
  15. Oct 22, 2009 #14

    tiny-tim

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    Got it!

    You used cos 31 radians instead of cos 31º :rolleyes:

    Everything perfect apart from that! :biggrin:
     
  16. Oct 22, 2009 #15
    I'm never using google calculator instead of my own again... :P Good catch, never would've though of that.
     
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