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Derivations of KE=1/2mv^2 and PE=mgh

  1. Jan 7, 2013 #1
    1. The problem statement, all variables and given/known data
    I need the derivations of gravitational potential energy and kinetic energy (without using calculus)

    2. Relevant equations
    PE = mgh \\
    KE = {1 \over 2}mv^2

    3. The attempt at a solution
    Potential Energy:

    Work done = [itex]Fs[/itex]
    But... [itex]F=ma[/itex]
    So substituting that in becomes [itex] mas [/itex]
    where a is the acceleration due to gravity (the g in mgh), and s is the height in mgh.

    Kinetic Energy (this is I am a lot less sure on):

    Work done = [itex]Fs[/itex]
    But... [itex]F=ma[/itex]
    So substituting that in becomes [itex] mas [/itex]
    And when u=0 [itex]v^2 = 0 + 2as[/itex]
    Which can be arranged to [itex]a = {v^2 \over 2s}[/itex]
    So substituting that in becomes [itex] = m \times {v^2 \over 2s} \times s[/itex]
    Which can apparently (so I was told) be rearranged to become [itex]{1 \over 2}mv^2[/itex]

    Can anyone let me know if these are correct, if not any advice is appreciated and if they are correct, any advice on how to word it correctly is also appreciated.

  2. jcsd
  3. Jan 7, 2013 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    It is fine but your explanation could be clearer. Start with the definition of Energy as "the ability to do work". The amount of work a body of mass m will do in falling from point A to B through a vertical distance h is a measure of its potential energy at A relative to B. That amount of work is the force ( = mg) multiplied by the distance through which the force acts (h).

    This you need a bit of help on. Start with the same definition of energy. The kinetic energy is the ability to do work due to a body's motion rather than its position.

    How much work will a body travelling at velocity v do before it stops? That is a measure of its kinetic energy.


    W = ΔKE = FΔs
    FΔt = -mΔv = m(v-0) = mv
    Δs = vavgΔt

    Last edited: Jan 7, 2013
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