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Conservation of Energy With Friction

  1. Nov 20, 2016 #1
    1. The problem statement, all variables and given/known data
    A 10 kg initially at rest is pulled 13m across a floor by a 50N force. if friction does 380J of work over this distance, what is the block's final velocity? http://imgur.com/a/zM1MX

    2. Relevant equations
    W=Fd, Ek=1/2mv^2

    3. The attempt at a solution
    Since the block was at initially at rest the Energy initially is 0, so I set up the equation like this 0 = Ek after + Efriction. Since friction is 380J i moved it to the other side, then did 1/2mv^2 = 380J to find V which I got 8.72 m/s, but the answer is 7.3m/s. Did I set up the equation wrong?
     
  2. jcsd
  3. Nov 20, 2016 #2

    kuruman

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    You forgot to include the 50 N force in your calculation.
     
  4. Nov 20, 2016 #3
    How would the entire equation look like in terms of Energy like Ek before = Ek after - E friction?
     
  5. Nov 20, 2016 #4

    kuruman

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    It is the work-energy theorem.

    KEfinal - KEinitial = Wfriction + W50 N force

    You need to find expressions for each one of the terms and put it together.
     
  6. Nov 20, 2016 #5
    Oh okay thank you. Last question is work and energy the same thing, and are they both scalar quantities?
     
  7. Nov 20, 2016 #6

    haruspex

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    Yes, they're both scalars. Energy is the more generic term, while work is used for energy transfer.
    In thermodynamics, work refers to energy transfer in mechanical form at the macroscopic level, while heat refers to energy transfer at the level of uncoordinated motion of molecules. Exactly where one draws the line is not defined, and may depend on context.
    In regard to this thread, note that in kuruman's equation both the work terms on the right refer to work done on the object. They will not both be positive.
     
  8. Nov 20, 2016 #7
    http://imgur.com/a/VHZyU For a question like this, does it matter which direction the velocity is? would I take the resultant velocity to find kinetic energy, so 1/2(m)(500m/s)^2? Or do I need to break it in to x and y components?
     
  9. Nov 20, 2016 #8

    haruspex

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    I assume you mean for the initial KE. Yes, try that.
     
  10. Nov 21, 2016 #9
    Sorry last question, about kuruman's equation, why won't both be positive since they're both scalars I thought direction wouldn't affect it?
     
  11. Nov 21, 2016 #10

    jbriggs444

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    Scalars can be positive or negative. The direction of the vector quantities that go into calculating a scalar may (or may not) affect the sign of the scalar. So the general answer is "it depends".

    For kinetic energy, ##E=\frac{1}{2}mv^2##, it turns out that one is taking the scalar product of a vector ##\vec v## with itself and multiplying the result by ##\frac{1}{2}m##. Direction does not matter. But in the case at hand, I think you are asking not about kinetic energy, but about work.

    We are considering the work done by friction and the work done by the 50N force. Is the work done by friction on the 10 kg object positive or negative? Is the work done by the 50N force on the 10 kg object positive or negative?
     
  12. Nov 21, 2016 #11

    kuruman

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    And remember, the work done by a constant force on an object is the product of three quantities, the magnitude of the force doing the work, the magnitude of the displacement of the object and the cosine of the angle between the force and the displacement. So, under what circumstances is the work done by the force on the object positive, zero or negative?
     
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