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Kinetic/Potential Energy - What am I doing wrong?

  1. Oct 13, 2016 #1
    1. The problem statement, all variables and given/known data
    What am I doing wrong here?

    Given the following two formulas for Kinectic Energy (KE) and Potential Energy (PE):

    PE = mgh
    KE = 1/2mv2

    If a 10KG weight was lifted 10 meters above sea level - I calculate its PE as:

    PE = 10kg x 9.81m/s x 10meters = 981 Joules

    If I let the ball drop I calculate its kinetic energy as (assuming nothing lost due to friction etc):

    KE = 1/2 x 10KG x 9.81m/s x 9.81m/s = 481 Joules

    However, in this case, I understood the KE should equal the PE?

    Which calculation have I got wrong?

    2. Relevant equations
    PE = mgh
    KE = 1/2mv2

    3. The attempt at a solution

    PE = 10kg x 9.81m/s x 10meters = 981 Joules

    KE = 1/2 x 10KG x 9.81m/s x 9.81m/s = 481 Joules
     
  2. jcsd
  3. Oct 13, 2016 #2

    haruspex

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    I presume you mean 9.81 m/s2.
    What makes you think the final velocity is 9.81m/s?
     
  4. Oct 14, 2016 #3
    I'm thinking this is the amount of time it takes to falls in 1 second and therefore the speed.
     
  5. Oct 14, 2016 #4

    CWatters

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    Take a look at the equations of motion..eg...

    V^2=U^2 + 2aS
     
  6. Oct 14, 2016 #5

    Simon Bridge

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    You just said that 9.8m/s is the amount of time it takes to fall in 1 second.
    Surely the amount of time it takes to fall in 1 second is... 1 second?

    Perhaps you mean that it falls 9.8m in the first second, so the average speed over that time is 9.8m/s ... but that is not the case either.
    It actually falls 4.9m in the first second.

    Have you heard of the kinematic equations (sometimes called suvat equations) and velocity-time diagrams?
    Once you have those, you can use the correct one (CWatters tells you which one above) and then do the kinetic energy calculation symbolically.
     
  7. Oct 14, 2016 #6
    Thanks so much Cwatters/Simon - a great help :-)

    -- SUVAT Equations --

    I know the following:

    (s) displacement = 10 meters
    (g) gravity = 9.81 m/s
    (t) time = 1 second
    (u) initial velocity = 0

    I can therefore use:

    v2 = u2 + 2as
    v2 = 0x2 + 2 x 9.81m/s x 10 meters = 196.2
    v2 = 196.2
    v = 14.007
    ----

    Back to my original question:

    KE = 1/2mv2

    KE = 1/2 x 10KG x 14.007 x 14.007 = 981 Joules

    This ties up with the PE

    :-)
     
  8. Oct 14, 2016 #7

    CWatters

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    In the end you didn't use "time = 1 second" to get the right answer but you might like to check if it really does take 1 second to fall 10m.

    Perhaps by using another SUVAT equation... v = u+at and solving for t.
     
  9. Oct 14, 2016 #8

    Simon Bridge

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    You did it numerically and showed that it was consistent for one example. maybe that was just a coincidence and you'd get a different result from chosing different numbers? This is why it is better to do the algebra using symbols ...

    potential energy lost falling h is ##U=mgh## (close to the surface of a spherical mass with gravitational acceleration at the surface of g)
    kinetic energy gained is ##K = \frac{1}{2}mv^2##

    from suvat equations, ##v^2=2as## ... in this case, ##a=g## and ##s=h## so:
    ##K=\frac{1}{2}m(2gh) = mgh = U##

    ... the advantage of doing it this way is that it now does not matter which numbers you pick, the two equations (related by the laws of motion) are always going to be consistent in this way. notice that it does not even have to be the Earth causing the gravity.
     
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