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Simple force and work problem

  1. Apr 11, 2015 #1
    1. The problem statement, all variables and given/known data
    See attached image


    2. Relevant equations
    See attached image


    3. The attempt at a solution
    See attached image

    I simply want some guidance pertaining to if I did this problem correctly. Is the diagram correct? In addition, I have a specific question. We have the equation ##\vec{F} = \frac{P}{\vec{v}}##. My question is if this equation is true, then why isn't the force in the direction of the velocity? I though that if a vector was written in terms of another vector, then the first vector had to be in the direction of the second (e.g. ##\vec{v} = \frac{\mathrm{d} \vec{s}}{\mathrm{d} t}##, where velocity has to be in the direction of the displacement).
     

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  3. Apr 11, 2015 #2

    Simon Bridge

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    The equation is not true.
    You can derive the correct equation from the definitions of work and power.
     
  4. Apr 11, 2015 #3

    haruspex

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    You can't divide by a vector. The equation is ##\vec{F}.\vec{v} = {P}##. A dot product does not behave like scalar multiplication.
     
  5. Apr 11, 2015 #4

    NascentOxygen

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    This is taxing my mind-reading finesse twice over! :smile:

    It took me a while to figure out what the examiner was talking about exactly, and I'm still not 100% sure I have it right.

    First, could you explain, in words, what you think the vector diagram should show?

    Meanwhile, 160/4 ≠ 80
     
  6. Apr 11, 2015 #5

    haruspex

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    Hint: write the F vector as the sum of its x and y components and substitute in the correct version of the power equation.
     
  7. Apr 11, 2015 #6
    The vector diagram should show the forces acting on the flat rock. There are three forces: the applied stick force, gravity, and the normal force. The normal and gravity cancel out, so the net force is that of stick, and since the question states that the stone is moving in the horizontal direction, it is specifically the x-component of the applied force that causes the acceleration. This is my reasoning. Anyhow, what exactly am I doing wrong? I'm not sure what to change because nobody has told me what I am doing incorrectly...

    Also, you're right that 160/4 does not equal 80, but I evaluated the force again, correctly I think, and it still turned out to be 80 N.
     
  8. Apr 11, 2015 #7

    Simon Bridge

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    I told you the equation you used was wrong and haruspex told you this was bacause you cannot divide by a vector... it is difficult to go into details more without doing the problem for you.

    Basically:
    You have not converted the vector equation to a magnitude equation correctly.
    Check the maths definition of the dot product, or the physics definition of work and power.
     
    Last edited: Apr 11, 2015
  9. Apr 11, 2015 #8
    So would this be correct?

    ##P = \frac {W}{t} = \vec{F}\cdot \vec{v} = Fv\cos \theta##
    ##F = \frac{P}{v\cos \theta} = \frac{160}{4.00\cos 60^{\circ}} = 80 N##

    Then

    ##W = F_{x}d = (80)(2) \cos 60^{\circ} = 80 J##

    Is this correct? If not, what am I doing wrong?
     
  10. Apr 11, 2015 #9

    Simon Bridge

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    That's what I'd have done from the same information.
     
  11. Apr 11, 2015 #10

    NascentOxygen

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    Your answer checks out. You'd calculate the same work if you said it's being moved at a steady speed for ½ second and her power is 160W.
     
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