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Power Question: Just Check My Work!

  1. Oct 19, 2006 #1
    A motorcycle (mass of cycle plus rider = 2.5 x 10^2 kg) is traveling at a steady speed of 20.0 m/s. The force of air resistance acting on the cycle and rider is 2.00 x 10^2 N. Find the power necessary to sustain this speed if the road is sloped UPWARD 37.0 degrees with respect to the horizontal.

    OK this is how im doing this problem (Check my work on here)

    First Step (finding net force):
    Fnet = Wsin37 + Fa (exerted) = (250)(9.8)sin37 + 200 = 1674 N (briefly, Just tell me if it's sin or cos here! -- main confusion)

    Second Step (finding power from givens):
    P = (Fcos37) V = (1674cos37) (20) = 2.67 x 10^4 W
    Again, Just check my work to see if i am doing it correctly. If not, then please give me a little help -- hint: im bad with conceptual understanding (try to help me visual what i did wrong!). Thanks and sorry if im asking for too much!
     
  2. jcsd
  3. Oct 19, 2006 #2

    Andrew Mason

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    The work done by the force is [itex]F\cdot s[/itex]. The work done by the air resistance and gravity (ie. the energy lost by the bike and rider) per unit time is

    [tex]\frac{d}{dt}W = \frac{d}{dt}F\cdot s[/tex]

    where s is the distance travelled and F is the vector sum of the forces on the bike. Since F is constant,

    [tex]\frac{d}{dt}W = F\cdot\frac{ds}{dt}[/tex]

    Since the bike travels at a constant speed (ds/dt is constant - kinetic energy loss is 0) what is the relationship between the rate of kinetic energy loss and the power that must be supplied to the bike?

    [hint: you have to determine the components of force in the direction of motion.]

    AM
     
    Last edited: Oct 19, 2006
  4. Oct 19, 2006 #3
    ummmm kinda lost! So should i find the distance instead of plugging the Velocity in the Power Equation (P = Work)??
     
  5. Oct 19, 2006 #4

    Doc Al

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    Looks good to me. The applied force (exerted by friction on the tires as a result of the engine driving the wheels) must equal that sum for the velocity to be constant.

    As Andrew said, you need to determine the force in the direction of the velocity. (Hint: You already did! Which direction is the applied force you found in step 1? Which direction is the velocity?)

    (Note: In the generic formula, P = F*V*cos(theta), theta is the angle between F and V. Don't just pick any old angle. What's the angle between F and V in this situation?)
     
  6. Oct 19, 2006 #5

    Andrew Mason

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    Your approach in the first part is correct but your analysis is a little unclear. What is the direction of this force?

    In the second part, you are using the factor cos37 without explanation. Why? To find the power, you simply multiply the component of total force in the direction of motion x the speed. Or you take the dot product of the total force and the velocity.

    AM
     
  7. Oct 19, 2006 #6
    Ok, is the angle between F and V 53 degrees (90-37 = 53) ? seems right??? :smile: :confused:
     
  8. Oct 19, 2006 #7

    Doc Al

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    Nope. Answer these questions (draw yourself a picture):
    -- The applied force needed to keep the motorcycle moving (which you calculated) acts in what direction?
    -- What direction does the motorcyle move?

    Once you've drawn these vectors on a diagram showing the inclined road, then you can determine the angle between them.
     
  9. Oct 19, 2006 #8
    ok it goes along the gravity sooo is it sin37 ??:smile:
     
  10. Oct 19, 2006 #9

    Doc Al

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    Enough with the guessing! Answer my questions. :grumpy:
     
  11. Oct 19, 2006 #10
    lol i am not guessing...i dont think im competant enough to visualize or draw out this scenario!!! The motorcycle is going up a 37 degree slope and a force exerted (by calculation) is 1674N. This is pretty much all i know...i dont know wth is between those angles (F and V). How do you even visualize that?? my thought is that they're on the same axis...so i dont know at all. Please HEEEELP me--an answer wont hurt will it?? :smile:
     
  12. Oct 19, 2006 #11

    Doc Al

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    Draw a simple diagram of an incline at 37 degrees with a blob on it representing the motorcycle...

    Come on... give it a shot. Let's see... if the motorcycle is going up the ramp, where would its velocity vector be pointing? Hmm... maybe... straight up in the air? :uhh: no... that can't be right... What do you think?

    Same for the force. Which way does the force have to push on the motorcycle?

    It's easier than you think! :wink:
     
  13. Oct 19, 2006 #12
    I think your notation is a little screwy....

    you wrote:
    Fnet = Wsin37 + Fa (exerted) = (250)(9.8)sin37 + 200 = 1674 N

    However, Fnet = 0! Fa = 1674.

    If you're moving at constant velocity then your net force must be ZERO.

    it should be, sum of forces = 0N = Fa(exerted) - 2E2 - mgcos(53). Solving for Fa gives Fa = 1675. You had the right number, but you labeled the number as Fnet, which is wrong.

    Do you see why I subtracted mgcos(53)? You need to project the weight of the bike in the direction parallel to the ramp surface. Sit down and show yourself why the projection of the bike's weight in the direction parallel to the ramp surface is mgcos(53). In this case mgsin(37) = mgcos(53).

    Now, to determine the power you need to expend you need to think about the definition of power. It is the rate at which work is done, or equivalently, the rate at which energy is expended or transferred. Work is force times distance in the case where force is constant over the distance of exertion. Technically, the net work on the system is ZERO, but you want to determine the work the bike needs to do for the net work to be zero. The work the bike does is Fa*d. Power = (Fa*d)/time, or, equivalently, Power = Fa*V, where V is the velocity of the bike....

    So, the power the bike needs to produce to cause an equilibrium situation so that no net work is done (and therefore no change in velocity) is simply Power = Fa*V = (1675)*(20) = 33,500 Watts, or 33.5kW.
     
  14. Oct 19, 2006 #13
    OK well i drew one way before i posted this question...

    Well, velocity would be in the positive direction going UP

    and

    the Force is the upward as well (maybe 1674N by itself with cos 180 = 0 ? -- NOT A GUESS)
     
  15. Oct 19, 2006 #14
    Hey Thanks for the explanation!! i'll study this one more in depth
     
  16. Oct 19, 2006 #15

    Doc Al

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    You got it (almost): The velocity is up the ramp; the force is up the ramp. The angle between them: 0 degrees (not 180)--they are in the same direction! cos(0) = 1.
     
  17. Oct 19, 2006 #16
    yo i appreciate your patience and help!! i was getting a little frustrated earlier errrrrrr, but it's all good now THANKS!! :smile:
     
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