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Introductory Power/Work Problem

  1. Nov 12, 2009 #1

    jacksonpeeble

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    1. The problem statement, all variables and given/known data
    1. Some gliders are launched from the ground by means of a winch, which rapidly reels in a towing cable attached to the glider. What average power must the winch supply in order to accelerate a 188-kg ultra-light glider from rest to 24.0 m/s over a horizontal distance of 51.0 m? Assume that friction and air resistance are negligible, and that the tension in the winch cable is constant.

    2. A person pushes a 16.0-kg shopping cart at a constant velocity for a distance of 28.0 m. She pushes in a direction 24.0° below the horizontal. A 32.0-N frictional force opposes the motion of the cart.
    a. What is the magnitude of the force that the shopper exerts?
    b. Determine the work done by the pushing force.
    c. Determine the work done by the frictional force.
    d. Determine the work done by the gravitational force.


    2. Relevant equations
    P=W/T
    W=F*D


    3. The attempt at a solution
    1. P=((188kg*24m/s)*51m)/(51m/24m/s)

    2. a. I assumed this would just be 16*28*cos(24), but this doesn't work. Once I have this, I can solve for B and probably C.
    b.
    c.
    d. w=0
     
    Last edited: Nov 12, 2009
  2. jcsd
  3. Nov 12, 2009 #2

    jacksonpeeble

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    Any help would be greatly appreciated!
     
  4. Nov 12, 2009 #3

    Redbelly98

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    If you're saying that T=D/vfinal, the problem with that is the glider does not have v=constant=vfinal during the entire time.

    How about using T=D/vaverage, assuming constant acceleration?

    Also, force F is not m vfinal. How about using the change in the glider's energy to find the work done on it?

    Regards,

    RB

    p.s. I'd like to suggest posting your 2nd question as a separate thread.
     
  5. Nov 12, 2009 #4

    jacksonpeeble

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    Thank you very much for the reply! Since the initial velocity is 0, I believe the average velocity would be simply .5*24=12. Correct me if I'm wrong, and please excuse my clumsy mistake! :-)

    Therefore 51m/12m/s=4.25s.

    (188*[change in energy])*51/4.25=p

    However, I still do not understand how to find the change in energy. A little clarification would be greatly appreciated. I'm probably missing something obvious, but I'm really drawing a blank.

    Thanks again!
     
  6. Nov 12, 2009 #5

    jacksonpeeble

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    Shoot, that's average acceleration, isn't it...

    I'm not on my game tonight. I really need help, though, the assignment is due tonight at midnight (online class).
     
  7. Nov 12, 2009 #6

    jacksonpeeble

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    Never mind...

    12739.8=(24/(51/12))*188*51/(51/12)
     
  8. Nov 12, 2009 #7

    Redbelly98

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    Correct!

    Yes.

    See below.

    Looks good. a*m*D / T = F*D/T = W/T = P :smile: I didn't check the arithmetic, but things are set up correctly on the right hand side.

    To do this using energy, realize that the work done W can be converted to either kinetic or potential energy, or a combination of both. So if you can calculate the change in total energy,

    ΔKE + ΔPE​

    you could equate that result with the work W. This should give the same answer as a*m*D.
     
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