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Homework Help: Work Forrmulas

  1. Jul 21, 2015 #1
    1. The problem statement, all variables and given/known data

    Guys, consider the above image.
    Let the incline to be smooth.

    If I want to find NET WORK ..

    There are 2 formulas ..

    W = Fd and W = Change in Mecahnical Energy

    In First Formula

    Wnet = Wf + Wmg
    Wnet = Fd - mgsin37 * d

    In second Formula

    Wnet = Change in KE + Change in PE
    Fd = Change in KE + Change in PE

    Not I got confused .. I know I am doing right..
    But in first formula: Wnet = Wf + Wmg but in the second I substittued Fd only for Wnet

    So which one is Wnet??? I hope you get my point

    2. Relevant equations
    W = Fd
    W = Change in KE + Change in PE

    3. The attempt at a solution
    not a problem needing solution. It is about the concepts
  2. jcsd
  3. Jul 21, 2015 #2
    I think my question was not clear enough..
    I will try to paraphrase it..
    I know that work is the vector product of displacement and force.
    If we have more than one force, the net work is the sum of work for each force.
    So, as we have in the incline: Wnet = Wmg + WF

    Now, we move to concept of work and its relation to energy.
    Net work is the change in the mechanical energy.
    Wnet = Change in KE + Change in PE
    Here, Wnet is only WF right? Not Wmg + Wf.

    So, sometime we consider the Wmg in Wnet and sometime not .. Please help me to resolve this confusion
  4. Jul 21, 2015 #3


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    Staff: Mentor

    If the incline is not frictionless, you need to include the work done by the force of friction. You cannot just drop it out.
  5. Jul 21, 2015 #4
    Thanks for your reply but I stated in the second line of my post that the incline is smooth, so no friction
  6. Jul 21, 2015 #5


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    Staff: Mentor

    Oh, well then the friction term goes away for both formulas, and they are the same, no? Maybe I'm missing your question... :smile:
  7. Jul 21, 2015 #6
    My confusion point is in the formula:

    Wnet = Change in Mechanical Energy
    What does Wnet mean here?
  8. Jul 21, 2015 #7


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    Staff: Mentor

    Just the change in potential energy ΔPE = mgΔh.
  9. Jul 21, 2015 #8
    I think you're confused because the Wnet that you use in the two different equations mean different things. In your first equation, the work is the sum of the work due to gravity and the work due to the applied force, which equals Wnet. An important detail that you're missing is that this Wnet in this case is also equal to the change in kinetic energy, meaning that you get the equation Change in kinetic energy = Fd - mgsin37d. In your second equation, work is considered to be only the work that isn't conserved. The change in energy on the right side now includes the change in kinetic energy. Since the change in PE that you use here is also equal to mgsin37d, you end up getting the equation Change in kinetic energy = Fd - mgsin37d, which is also what you get in the first equation if you sub in Change in kinetic energy for Wnet. Basically, you accounted for kinetic energy in the work in the second equation but not the first. They are both right, but the two values for work aren't referring to the same thing in both cases.
  10. Jul 21, 2015 #9
    Thanks, I figured out that they are equal.
    But if the questions asked me to find Wnet .. Which one is Wnet?

    Lets take this question:
    A 1 kg rock is thrown directy upward with an initial velocity of 20 m/s. Assume no air resistance, the net work done to reach the maximum height is: ___

    Now ..

    If I used the first formula .. The only force I have is force of gravity. So Wnet = Wmg = mgd = 1*10*d = 10*d = 10 * 20 = 200 J
    But by using second formula .. Wnet = Change in Mechanical Energy = 0 J
  11. Jul 21, 2015 #10
    I read the replies again ..
    Wnet = W1+ W2 + W3
    This Wnet = KE NOT Wnet = KE + PE
  12. Jul 21, 2015 #11


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    Science Advisor
    Homework Helper
    Gold Member

    It depends whether you mean the net work done by the force (##F d = \Delta PE + \Delta KE##), or the net work done on the block (##\Delta KE=F d - mgh##).
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