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Problem with f=ma and E=0.5mv^2

  1. Mar 24, 2012 #1


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    EDIT: The question has been answered, and I don't need help anymore. Thanks a lot everybody!

    Hi. This is something that has been puzzling me for a while, probably just due to my own stupidity. It seems to me that this should be really simple, but I don't know what I'm doing wrong though, so now I'm putting it out here.

    Imagine an object with a mass of 1 kg being accelerated at a rate of 1 m/s^2 over 10 seconds.
    We can then calculate the force being applied to the object
    F=1*1=1 N

    We can also calculate how far the object has moved over these 10 seconds

    According to Wikipedia, the work (W) is the product of a force times the distance through which is acts, measured in Joules, so
    W=F*d=m*a*d=55 J

    The velocity v at which the object moves is t*a=10 m/s
    Then we can calculate the kinetic energy E

    E=0.5*m*v^2=50 J

    But shouldn't E and W be the same?
    For example, if we take the object, now with the kinetic energy E and make it go up a hill on a planet where the gravitational acceleration is 1 1m/s^2 (which is =a from before, of course) we will convert the kinetic energy to potential energy, and we can calculate how far up the hill the object will go, as E(kin)+E(pot) is constant

    h=50 m

    But shouldn't h be 55 metres, as that was the distance the object moved when we accelerated it?
    I hope you understood what I meant, I'm absolutely horrible at explaining my thoughts.
    Thanks in advance! :)
    Last edited: Mar 24, 2012
  2. jcsd
  3. Mar 24, 2012 #2
    The distance travelled is not 55m.
    Check your calculation
  4. Mar 24, 2012 #3


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    Distance travelled over time

    after 1s: 1 m (velocity 1 m/s)
    2s: 3 m (velocity 2 m/s)
    3s: 6 m (velocity 3 m/s)
    4s: 10 m (velocity 4 m/s)
    5s: 15 m (velocity 5 m/s)
    6s: 21 m (velocity 6 m/s)
    7s: 28 m (velocity 7 m/s)
    8s: 36 m (velocity 8 m/s)
    9s: 45 m (velocity 9 m/s)
    10s: 55 m (velocity 10 m/s)
  5. Mar 24, 2012 #4

    D H

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    Staff Emeritus
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    Your calculations are incorrect. Given a constant acceleration a, distance d as a function of time is given by [itex]d=\frac 1 2 a t^2[/itex].
  6. Mar 24, 2012 #5


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    Ah, well that obviously solves it. Thanks a lot.
  7. Mar 24, 2012 #6
  8. Mar 24, 2012 #7


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