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Work Done by Constant Force

  1. Dec 4, 2012 #1
    1. The problem statement, all variables and given/known data

    lyY5C.jpg

    2. Relevant equations

    W = fdcos∅

    3. The attempt at a solution

    I know how to solve the problem using potential energy (i.e. mgh) and the answer is 4.41 J.

    My question is, how do I solve it using the work done by the applied force using fdcos∅?

    The work done parallel to the incline is 82sin(53) = 65.49
    The distance the the box moves up the incline when h = .15 is (I think), .15cos(53) = .09027.

    However, when multiplying them together (.09027*65.49) I get 5.9 J.

    What am I doing wrong? I'm pretty sure it's in finding the distance that the box travels up the incline.
    -edit- I see what I'm doing wrong and it is in the distance the box travels up. I need the angle of the incline, not the force to ground. Is there any way to find that angle?
     
    Last edited: Dec 4, 2012
  2. jcsd
  3. Dec 4, 2012 #2

    PhanthomJay

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    no, the angle of the incline is not given and cannot be calculated . Does it matter?
     
  4. Dec 4, 2012 #3
    No, it's fine. I was just wondering I guess. Thank you :)
     
  5. Dec 4, 2012 #4

    PhanthomJay

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    Well wait a second, the comp of the force parallel to the incline must be equal to the comp of the weight parallel to the incline. That is, with some trig, mgsinψ = Fsin(53-ψ). Or, mg/F = sin(53-ψ)/sinψ. Here, ψ is the angle of the incline. So cranking out the numbers, 3(9.8)/82 = 0.36 = sin(53-ψ)/sinψ. i would think you can solve this for ψ , but how??? I don't know!
     
  6. Dec 5, 2012 #5
    I think there's 2 ways of doing this:

    (1) The change in KE is 0, so the total work done on the box is zero.
    So the work done by F is the negative of the work done by gravity.
    W_gravity = -mgh ( h is the vertical height gained by the box ).
    So
    W_force = mgh = 4.41J

    (2) Put x-axis along the ramp pointing up.
    Then
    F_net_x = Fcosα - wsinθ = 0
    ( α is angle F makes with the x-axis, and θ is the angle the incline makes with the horizontal surface ).
    So, Fcosα = wsinθ
    So, W_force = Fdcosα, (where d is the distance the box travels along the incline ).
    → W_force = Fdwsinθ/F = dwsinθ
    Using some trig, d = h/sinθ
    →W_force = hwsinθ/sinθ = hw = mgh = 4.41J
     
  7. Dec 6, 2012 #6

    PhanthomJay

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    Oh yes, sorry, i was focusing on determining the angle of the incline instead. The angle that the incline makes with the horizontal is of course____?____ degrees. ???
     
  8. Dec 6, 2012 #7
    As you've already pointed out :
    mgsinψ = Fsin(53-ψ)
    Expand out the sin(53-ψ), divide through by sinψ, leaving the only unknown as tanψ, leading to ψ = 39.7°
     
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