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2 Hard Problems Homework

  1. Oct 10, 2005 #1
    A competition-cyclist closing the goal. Towards the end-rush she raises her speed from 12m/s to 20m/s on 6.5s. The Cyclist and the cycles masses are together 70kg. Calculate acceleration The accleration is 1,23 no problem i calculated that but the next question is hard.

    The Cyclist pushes forward a power of 150N. How much power is pushing backwards. Which types of powers are there?
    How shall i Calculate the power backwards here?

    To the second problem....

    A person who weights 76,5kg hangs up in a helicopter with a line. Which powers utilizes the person and how big are theses powers.

    a) In the beggining when the acceleration is 1,3m/s^2
    b)When the speed is constant with 2.0m/s
    c)On the top, where the acceleration is -1,3m/s^2

    How do i calulate this? My mind stands still... please help me

    Thanks
    Izekid
     
  2. jcsd
  3. Oct 10, 2005 #2

    Päällikkö

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    F = ma, where F is the net force. Is this enough help?
     
  4. Oct 10, 2005 #3
    Nope in the cyclist problem you shall calculate with Friciton and Air-resistance.
    And the result backwards shall be 64N.


    The second problem I resolved by myself :smile: :smile: :smile: :smile: :smile:
    Because 76,5 * 9,82 = 751,23 N
    and 76,5 * 1,3 = 99,45 N = 850,68N =0,85kN
    And so on (I forgot to take the gravitionconstant)

    But PLZ help me with the cyclist... :yuck: :yuck: :yuck:
     
  5. Oct 10, 2005 #4

    Päällikkö

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    As the forces added must result ma:
    [tex]F_f + (-F_b) = ma[/tex]
    f stands for forward, and b for backward.
     
  6. Oct 10, 2005 #5

    VietDao29

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    ???
    Sorry, I just wonder what kind of help is that?
    I wouldn't use power here... I'd use force instead.
    So you know that the cyclist has an acceleration of 1.23 m /s2. You also know his mass (his own mass, and the bicycle mass), so you will be able to calculate the net force acting on him, right?
    Fnet = ma = ...
    Then you will have Fnet < 150 N, right?
    He's pulling himself forward with a force of 150 N, but the net force on him is less than 150 N, so there must be some force that slow him down, ie it acts in the oposite direction. Can you find the magnitude of that force?

    For #2, you seem to forget #c :wink:
    The net force for c will be:
    Fnet = ma = 1.3m.
    Remember that he's decelerating, so the net force will point down (since he's moving upwards).
    Viet Dao,
     
    Last edited: Oct 10, 2005
  7. Oct 10, 2005 #6
    thx that did it
     
  8. Oct 10, 2005 #7

    Päällikkö

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    Granted, it wasn't much help. I'd call it more like a reminder. In a problem like this, there isn't mighty lot one can say without solving the problem. I didn't want to give away the answer, as I personally hate being guided too much.
     
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