1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Formulas converting Potential Energy to Kinetic Energy

  1. Mar 16, 2014 #1


    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data

    A 5kg ball (m), is to be dropped 1 meter (h) in a vacuum with no resistance in the earth's gravity (g).

    Use the speed and motion equations to show that its KE at impact is equivalent to its PE at its starting point.

    2. Relevant equations

    PE = mgh
    KE= 1/2mv^2
    S = vt + 1/2gt^2
    S = vt

    3. The attempt at a solution

    Since h = 1
    PE = mg

    To calculate KE, we need to calculate the time taken to fall 1 meter and hence calculate the velocity.

    h = 0 + 1/2gt^2
    t^2 = 2/g

    S = vt
    v = h/t
    v = √g/2
    v^2 = g/2

    KE = 1/2mv^2
    KE = 1/2 x mg/2
    KE = mg/4

    I've now done this a few times and by the above calculations, the KE is 1/4 of the PE.

    There has to be an error somewhere in my calculations, hopefully someone help enlighten me :)
  2. jcsd
  3. Mar 16, 2014 #2
    Try using (vt)2 = (v0)2 + 2aΔx. Your 1/4 problem should be gone then :cool:
  4. Mar 16, 2014 #3


    User Avatar
    Science Advisor
    Homework Helper
    2017 Award

    I remember something like s = s0 + v0t+ 1/2 a t2
    And in this case s0 is 1 m, v0 = 0 and a = -g

    both OK (if interpreted somewhat liberally)

    "S = vt" is NOT ok, though. This is only if v is constant, and v is not constant: there is a constant acceleration (namely -g)

    "v = h/t" idem, and the "therefore v = √g/2 and v^2 = g/2" needs to be reviewed.

    (What you want here is something like v = v0 + a t with v0 = 0 and a = -g )

    " KE = 1/2mv^2 " true again, now we only need a better v .
  5. Mar 16, 2014 #4


    User Avatar
    Gold Member

    Thanks BvU.

    v = at makes all the difference. You are correct, during the fall, a is constant, not v!

    What do you mean by (if interpreted somewhat liberally)?
  6. Mar 17, 2014 #5


    User Avatar
    Science Advisor
    Homework Helper
    2017 Award

    s = s0 + v0t+ 1/2 a t2 ; fill in
    (in a coordinate system where up is positive, so g = -9.81 m/s2):
    0 = h - 1/2 9.81 t2 bring to other side:
    1/2 9.81 m/s2 t2 = h

    The liberties are swapping s and s0 and the sign of g.
    You have the benefit of the doubt (it comes out all right this time, but is that by coincidence?):
    • I can see you take g = +9.81 m/s2, because you end up with t2=2h/g
      -- also missed a factor h there: h is 1, so the value isn't the issue but you need the dimension of length to get s2 --
    • but if you take g positive, you should really take h negative, )
    but mistakes with signs and dimensions are made very easily, so be careful. Better to work explicitly and conscientiously.
    Last edited: Mar 17, 2014
  7. Mar 17, 2014 #6
    Just a note - if you're solving for t2, then making h negative if you have gravity as positive is totally fine. However, if you're solving for t, just a hint that you can't take the root of a negative expression and that you can do something about it.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted