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Horse Power

  1. Jun 15, 2008 #1
    I have been working on this problem for awhile now and I cant get anywhere. Can anyone help?

    1) A horse pulls a cart with a force of 155.0 N at and angle of 32.0o with respect to the horizontal and moves along at a speed of 13.8 km/hr. How much work does the horse do in 8.6 min?

    2) What is the average power of the horse?
     
  2. jcsd
  3. Jun 15, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi qman316! Welcome to PF! :smile:

    (I assume it means that the road is horizontal, but the angle of the force from the horse is 32º)

    1) Do you know how to calculate work? :smile:
     
  4. Jun 15, 2008 #3
    First off, draw a picture. Once you have that think of what work and power really are. how do you find work?
    how do you find power?
    what equations do you use?

    i also think it may be better to figure out number 2 before number one. but that my be just me.
     
  5. Jun 15, 2008 #4
    All the equations I have to figure out Work involve knowing the distance, which I do not know in this problem. Is there another way to find work without knowing a distance?
     
  6. Jun 15, 2008 #5
    hmmm....well since the horse and cart are moving at a constant speed then the NET work is zero (the work the horse does and the work friction does cancel out resulting in no change in KE and thus no acceleration). However, the work the horse does is found by projecting the line of force caused by the horse onto the direction of motion. You would then multiply this by the speed to get the power (rate at which work is done). You can then multiply the power by the time (you need to convert minutes to hours though) to get work.
     
  7. Jun 15, 2008 #6
    not all the equations to find work requires distance. do you know what the formula for kinetic energy is?
     
  8. Jun 16, 2008 #7

    tiny-tim

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    Hi qman316! :smile:

    But you do know the distance …

    you know the time and the speed, so the distance is … ? :smile:
     
  9. Jun 16, 2008 #8
    how i would calculate it is to multiply the force from the horse by the distance travelled (which is just the velocity multiplied by the time)... F*d=Work

    am i missing something?

    the angle really has nothing to do with it unless your figuring out the work done by gravity right?
     
    Last edited: Jun 16, 2008
  10. Jun 16, 2008 #9

    tiny-tim

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    Hi shamrock5585! :smile:
    Yes … you're missing how horses and carts work! :biggrin:

    (how old is this book? :rolleyes:)

    oh … and this isn't your thread, so you won't give the answer, will you? :wink:
     
  11. Jun 16, 2008 #10
    so.... d=v*t which is 13.8km/hr(.1433hr), d=1.978 then F*d=W which is 155N*1.978=306.59, what did I do wrong cuz it says 306.59 is the wrong answer?
     
  12. Jun 16, 2008 #11

    HallsofIvy

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    The horse moves at a speed of 13.8 km/hr. Since your are asked for average power, not work, you can start by assuming any time you like- say 1 hour. How far does the horse pull the cart in one hour?
     
  13. Jun 16, 2008 #12
    the question in part a asks for the work and b the average power
     
  14. Jun 16, 2008 #13

    tiny-tim

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    Hi qman316! :smile:
    You left out the angle! :wink:
     
  15. Jun 16, 2008 #14

    Kurdt

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    If its still wrong after that check the units.
     
  16. Jun 16, 2008 #15
    ur answer should probably be in newton meters
     
  17. Jun 16, 2008 #16
    i was suggesting that the angle did not mean anything because i assumed the force was at an angle of 32 degrees instead of the force being horizontal and the box moving up a hill at angle 32 degrees...
     
  18. Jun 16, 2008 #17
    Either if the horse is walking on a slope of 32 degrees, or if the 'connection' from the horse to the cart is at an angle of 32 degrees, you still need to use it!


    I assume it is meant that the connection is not mounted horizontally between the horse and the cart. Then, the force that is given is the total force, at an angle of 32 degrees.

    What is the equation for work? I saw you use "W = Fd" before but that is wrong! Can you see from this what you did wrong?
     
    Last edited: Jun 16, 2008
  19. Jun 16, 2008 #18
    Let me just correct myself there (since I cannot seem to edit my post anymore).
    The equation 'W = Fd' is valid, but only under a certain condition, which has to do with the direction of the force.
     
  20. Jun 16, 2008 #19

    Kurdt

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  21. Jun 17, 2008 #20
    well wouldnt W=Fd still be valid but force would be Fcos32 now?

    to clarify what i meant before is that if the horse was walknig up the path the angle does not matter... the angle between force and displacement is 0 so Fcos0=F because cos0 =1

    so the horse is walking horizontal with a rope tied to a box pulling it up a slope of angle 32 degrees?
     
    Last edited: Jun 17, 2008
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