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Kinetic Energy and Work Homework Help

  1. Mar 12, 2013 #1
    1. The problem statement, all variables and given/known data
    A 50 kg mass is pulled at a constant speed up a 20° incline which is 12m long and 4.1m high.

    μk = 0.32

    a) find the work done by tension, T.
    b) find the work done by friction.
    c) find the work done against gravity.


    2. Relevant equations

    W = F x D
    Work = force x Distance

    W = Fd x cosθ


    3. The attempt at a solution

    I still confused on the concept of this.

    i know that you have to use Fd x cosθ, but i am confused on how to set it up.
     
    Last edited by a moderator: Mar 12, 2013
  2. jcsd
  3. Mar 12, 2013 #2

    berkeman

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    Staff: Mentor

    Always start this kind of problem with a free body diagram (FBD). Can you sketch one and post it as an attachment? And use the FBD to show the forces that you are being asked about in the problem...
     
  4. Mar 12, 2013 #3
    Frictional force is the constant of friction multiplied by the normal force, which is the force equal and opposite to the gravitational force.
     
  5. Mar 12, 2013 #4
    This is my FBD
     

    Attached Files:

  6. Mar 12, 2013 #5

    berkeman

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    Staff: Mentor

    Good, so now you can start writing equations for each of the works that you are asked to find...
     
  7. Mar 12, 2013 #6
    Im still confused on how to do that, all i know is that:

    f// = mgsinθ
    fn = mgcosθ
     
  8. Mar 13, 2013 #7

    vela

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    Staff Emeritus
    Science Advisor
    Homework Helper
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    What specifically is confusing you about applying the formula? You have three quantities, F, d, and θ. What do these variables represent?
     
  9. Mar 13, 2013 #8
    KTiaam - remember to put the frictional force in your free-body diagram.
     
  10. Mar 14, 2013 #9
    So i know how to find F of gravity and force perpendicular how do i find force friction?

    Since i know F perpendicular would it be just the opposite?

    F perpendicular is equal to 46.045
    i got that by:

    mass x gravity x cos 20°

    = 5 kg x 9.8 x cos 20°
    =46.045

    would force normal be negative then?
    or is force of gravity negative?

    i know that Force of friction is:

    coefficient of friction x force normal

    am i on the right track?
     
  11. Mar 15, 2013 #10
    Yes.
    For the direction of the frictional force - remember that it always opposes the motion.
     
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