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(Applied Force)⋅d= 1/2mv^2 versus (Net Force)⋅d= 1/2mv^2

  1. Aug 28, 2015 #1
    Suppose I have this object of 100kg and apply 1000N for 100 seconds with 900N of resistance force for the duration of the applied force.

    I calculated the distance to be 5000m.

    So the Work I do on the object is F*d = 1000N*5000m = 5 million J

    However, I calculated the final velocity to be 100m/s

    so the Kinetic Energy is 500,000 J

    and since [itex]\frac{1}{2}mv_{f}^{2} - \frac{1}{2}mv_{i}^{2} [/itex]

    That must mean Work must equal 500,000 J as well.

    The only way I can make F*d = 500,000 is if I use [itex]F_{net}[/itex]:

    [itex]F_{net}*d = 100N*5000m = 500,000 J [/itex]

    My question is, how do I refer to this type of work?

    Applied Force * Distance = Work I did on the object

    Net Force * Distance = What type of work? Or is it not referred to as work?
     
  2. jcsd
  3. Aug 28, 2015 #2

    BvU

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    You could call it useful work, since the efficiency of your activity is only 10%: 0.5 MJ is converted into kinetic enery and 4.5 MJ (900 N x 5000 m) is converted into (waste) heat.
     
  4. Aug 28, 2015 #3

    CWatters

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    It sounds like you forgot about the work done against the resistive force?

    The work you did (5,000,000J) breaks down as..

    4,500,000 joules against the resistive force (900 * 5000 = 4,500,000)
    and
    500,000 joules converted to KE
    total
    5,000,000 Joules

    What you call the 500,000 joules converted to KE is up to you. In some situation that might be the useful work. In others it might be the wasted energy. Depends on the situation. If this was a drag racing car you could argue both were useful.
     
  5. Aug 28, 2015 #4
    If I understand correctly, the work I did on the object is Force * distance in the direction of motion.
    The Work that the friction did on the object is Force * distance in the opposite direction.
    And the useful Work or KE (or the energy the box has gained that it can now apply to another object) is the net Force * distance.

    Like so:


    100N * 5000m = 500,000J
    _______
    l l
    900N * 5000m = 4,500,000J <------------ l l-----------> 1000N * 5000m = 5,000,000J
    l_______l


    Therefore, if I know the Force applied and the Force of resistance then finding kinetic energy is basically as easy as finding net force.

    Hope I understand correctly.
     
  6. Aug 28, 2015 #5
    The change in KE is always equal to the net work. This result is known as the "work-energy theorem". Maybe you look it up if you did not hear about it yet.
     
  7. Aug 29, 2015 #6
    Wow those dashes and l's did not come out how it originally looked in the text box when I wrote it. My apologies for that confusing mess!

    It was supposed to be an illustration of a box with force vectors * distance.

    Let me please restate that post entirely without the butchered illustration:

    ______________________________________________________________________________

    If I understand correctly, the work I did on the object is Force * distance in the direction of motion.
    The Work that the friction did on the object is Force * distance in the opposite direction.
    And the useful Work or KE (or the energy the box has gained that it can now apply to another object) is the net Force * distance.


    1000N * 5000m = 5,000,000J ---------------------> Work I apply to the object in the positive direction


    <------------------- 900N * 5000m = 4,500,000J Work the road applies to the object in the negative direction

    100N * 5000m = 500,000J ----------> Net amount of Work done on the object, which can also be calculated by Fd = 1/2 mv^2 = KE


    Therefore, if I know the Force applied, the Force of resistance, and the distance of an object's motion, then finding kinetic energy is a similar process to finding net force.

    _________________________________________________________________________________


    Does this make sense?
     
  8. Aug 29, 2015 #7
    Note: Please disregard post # 4 as I thought I'd be able to create a free body diagram using some dashes and letter l's, but it turned into a mess more than anything.

    Btw, Nasu,

    My physics book doesn't mention the term work-energy theorem. It just gives the equation fd=1/2 mv^2. It does not go in to detail to the degree of specifying a final and initial velocity. It just presents the equation as though all velocities start from 0. Then it goes straight into an example problem without any description. I just wanted to get a better sense of it which I am very appreciative of being able to seek further understanding here. Thank you.
     
    Last edited: Aug 29, 2015
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