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Homework Help: General question regarding positive and negatives in equations?

  1. Sep 29, 2013 #1
    1. The problem statement, all variables and given/known data

    I am trying to understand when the positive or negative of a specific value in an equation (work-energy and momentum-impulse) is to be evaluated based off the motion diagram and when it is not?

    2. Relevant equations

    The ones in particular I am unsure about is work-energy and impulse-momentum:
    PEi + KEi + W = PEf + KEf
    pi + J = pf

    3. The attempt at a solution

    I know with kinematics, you must always evaluate the values compared to the motion diagram and assigned positive/negative directions. I know the same goes for forces when using dynamics. I am mainly trying to make sure I understand it right when it comes to the work-energy and impulse-momentum equations.

    If I understand correctly, for PE, the g is always considered positive regardless of the motion diagram. For KE, I believe the same goes for the v value. Then for the W, I know the cosθ determines whether it will be positive or negative and the F and Δd are absolute values, making both positive (or 0 of course). So, if I understand correctly, the only thing that takes the positive/negative directions into consideration in work-energy is the cosθ, correct?

    As far as momentum-impulse, again if I understand correctly, in these equations the direction of the force and velocity will change each of their respective signs, correct?

    Our instructor has not been clear on this, threw both at us the same day, and then sometimes he uses the + or - from problems, sometimes he doesn't. After spending hours on my own trying problems, I think I have it figured out, but I would really appreciate someone confirming this for me.

    Also, for future reference, how can you tell when an equation evaluates certain values based on the motion diagram +/- values and when it is evaluated as a magnitude (abs. value) instead?
  2. jcsd
  3. Sep 30, 2013 #2


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    It's not that g is always considered positive, it's that g is to be multiplied by a downward displacement (corresponding to loss of PE) to a baseline, and g and the displacement must be measured in the same direction, both positive or both negative. Sorry, gotta go now... More later.
  4. Sep 30, 2013 #3
    So does that mean that if you changed g to a -g, you would have to change the h direction that causes a loss of height to be positive to coincide with the -g? Just trying to make sure I understand you correctly.
  5. Sep 30, 2013 #4


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    It depends what you mean by 'g'. If you mean 'the acceleration due to gravity' then it is just g, but its value is either 9.81 ms-2 or -9.81 ms-2, depending on whether you measure the vertical direction as positive downwards or upwards. Similarly the height h. So the PE is always -mgh. If an object mass M is 10m above the baseline and upwards is positive then h = +10m, g = -9.81 ms-2, the PE is -Mgh = +98.1M m2s-2. (Alternatively, you can regard g as standing for +9.81 ms-2 then say the accn due to gravity is +g or -g depending on whether down or up is positive. I'm not at all sure whether one of those two views is standard.)
    Similarly for the KE, it doesn't matter whether v is positive or negative, v2 is always going to be +ve.
    For force*distance, again, the signs matter. Distance and force are measured in the same direction, and work done by the force is F d. If you push on a block with force +3N but someone else pushes harder from the other side and you end up retreating 1m then the distance is -1m, so the work done by you is -3J.
    So in none of these cases is it true that we just take the absolute values. The sign comes out correctly by applying the rules.
    I'm not sure what you meant by the reference to cos theta wrt W. What you do have to watch out for with 'work done' is whether it's work done by (whatever) or work done on that. In my pushing example, the work done by you was -3J, which is the same as saying the work done on you was +3J.

    I hope this is clearer now.
  6. Sep 30, 2013 #5


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    When you set up the problem, you determine the "coordinate system"- that is, where things are "0" and which directions are positive and negative. It is common to take positive to be "up" and negative "down" but that is a convention.
  7. Sep 30, 2013 #6
    I get it now. Thanks for clarifying. After toying with some problems, changing my coordinate systems, and seeing what does and doesn't come up with the same values, I think I got it all down now. I appreciate you all taking the time to help me out with this. Take care.
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