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Knowing when to use mgh vs mg

  1. Mar 14, 2009 #1
    Can somebody please help me understand in which situations (for a intro physics 1 class) I would be required to use mg versus mgh? Neither my teacher nor my textbook have really gone into when you use which, rather they seem to arbitrarily use one, and I would like to understand why. Thanks for the help!
     
  2. jcsd
  3. Mar 14, 2009 #2

    Dale

    Staff: Mentor

    Look at the units. mg is in units of force (kg m/s²), while mgh is in units of energy (kg m²/s²).
     
  4. Mar 14, 2009 #3
    So do I use mgh when dealing with the energy principle and mg when using the momentum principle?
     
  5. Mar 14, 2009 #4

    Dale

    Staff: Mentor

    You do use mgh when dealing with energy.

    What are the units of momentum? Do they match the units of mg?
     
  6. Mar 14, 2009 #5
    Newton's second law states: F = m*a, in which a is the acceleration. In the case of gravitation, a = g = 9,1 m/s².

    Work = W = F*s.
    In case of gravitation, F = m*g, and s could be represented by h (height). The work done equals the change of potential energy. So potential energy U = W = mgh.
     
  7. Mar 14, 2009 #6
    i thought work was the change in kinetic energy
    and work is the integral of F*s
     
  8. Mar 14, 2009 #7
    1) Work can be the change in kinetic energy, but it doesn't have to be so. If a resultant force F would act on a car for a certain distance s, the work done would equal the change of kinetic energy: dT = W = Fs. But if you consider a weight that you want to lift up, you would need a force equal to the gravitational force (in opposite direction) to lift it to a certain height (h). In this case, the work done isn't converted into velocity (kinectic energy) but into height, thus potential energy.

    Another, maybe an easier way of looking at the problem is as follows. Consider an object of mass m at a certain height h. The gravitational force Fg=mg acts on the object. The object accelerates until it hits the earth (the mass will have velocity and thus kinetic energy). Because of conservation of energy, another ''kind'' of energy has lessened in order for the total change of energy to remain zero. This energy is the potential energy, and equals the work done by Fg.

    2) [tex]W=\int{F\cdot ds}[/tex]

    If F is constant, which is the case for gravitation (although not 100% accurate, since the force is less on a greater distance)

    [tex]W=\int{F\cdot ds = F \int{ds} = F\cdot s[/tex]
     
    Last edited: Mar 14, 2009
  9. Mar 14, 2009 #8
    how do u input the integral symbol in ur statement.
     
  10. Mar 14, 2009 #9
    Last edited by a moderator: Apr 24, 2017
  11. Mar 14, 2009 #10

    robphy

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    Science Advisor
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    Gold Member

    Work-done-by-a-force is the line-integral-of-F along a path.

    Using Newton's Second Law, the net-work (work done by the net-force) is the change-in-the-kinetic-energy.

    When there are conservative-forces doing work, one can define a potential energy function and then define that work as minus-the-change-in-potential-energy.
     
  12. Mar 14, 2009 #11
    yes i understood your explanation , thanks
     
  13. Mar 15, 2009 #12

    jtbell

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    Staff: Mentor

    The net work done by all forces acting on an object equals the change in the object's KE.
     
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