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When I do the math, GPE does not equal KE after an object drops.

  1. Oct 25, 2009 #1
    I teach 9th grade physical science, and we were doing an activity last week investigating whether GPE before an object falls = KE at the point the object hits the ground. This was a pretty crude activity involving dropping things out the window and timing with stop watches, so when the data didn’t look very good, I was not surprised. We were solving for velocity. Later, I got to wondering what the velocity should have been, so I calculated the velocity using the distance the object fell and g. Then when I calculated KE using a mass of 1 kg and PE using a weight of 9.8 N, the KE was ½ what it should have been----or the PE was twice what the KE was. I am no physicist, and I feel dazed and confused. Help please. We were dropping from ledge 6 meters off the ground.
     
  2. jcsd
  3. Oct 25, 2009 #2

    Doc Al

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    Staff: Mentor

    Please show the details of your calculation.
     
  4. Oct 25, 2009 #3
    At 6 meters, you should be getting about 58.8 Joules/Kg which would give a velocity of about 10.8 m/s. Are these the values you are getting/measuring?
     
  5. Oct 25, 2009 #4
    Okay. I set this up on Excel, and I may have messed up. I put in an equation first to solve for change in velocity considering acceleration due to gravity (9.81m/s2) x time. Then I made a column for time, which I kept changing until I got the distance I wanted (6.045m)using the equation distance = velocity x time. I know this is crude, but I don't have a physics book with an equation to solve for distance, so I thought this would work. But apparently not.

    I came up with 7.7 m/s for this distance of fall with a time of 0.78 seconds.
     
  6. Oct 25, 2009 #5

    Doc Al

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    That equation gives you the final speed after the given amount of time.
    Distance = velocity x time only works if the speed is constant or if you use the average velocity. Since you start from rest, the average speed is just half of your final speed. That's your problem.
     
  7. Oct 25, 2009 #6
    Thank you so much. This was making me crazy!
     
  8. Oct 25, 2009 #7
    Ok, I think I see your problem.

    You didn't use

    [tex]d=v_{i} t + \frac{a t^{2}}{2}[/tex]

    for your distance formula.

    Solving for t with no initial velocity gives

    [tex]t=\sqrt{\frac{2 d}{a}}[/tex]

    With a distance of 6 meters and an acceleration of 9.8 m/s^2 it gives a fall time of about 1.1 seconds. Plugging that into [tex]v=a t[/tex] will give a final velocity of about 10.8 m/s.

    [Edit]

    Yeah, what Doc Al said.
     
  9. Oct 25, 2009 #8
    To check that initial GPE equals final KE, there's no need to refer to time at all.
    The formula that gives the final speed is
    v_f ^2 = v_i ^2 + 2as.
    Dropping from rest, initial speed v_i = 0,
    v_f ^2 = 2as.
    v_f = sqrt [2as]
    a = 9.8 m/s^2.
    s = 6 m.
     
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