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Acceleration due to graviy

  1. Mar 21, 2009 #1
    1. The problem statement, all variables and given/known data
    Acceleration due to Graivty, dropping object 1 from recorded height for recorded time
    Δd = -2.36 (falling object)
    Δt = 0.86s
    vi = 0 (start from rest)
    a = ?

    Δd = Vi(Δt) + ½ a (Δt)^2

    2. Relevant equations
    I think I could use this equation, but I'm not sure what Vf would be,

    a= Vf - Vi / Δt

    I remmeber my teacher saying that Vf would be 9.8m/s DOWN, because of gravity but 9.8 is supposed to be an acceleration.

    3. The attempt at a solution
    Δd = Vi(Δt) + ½ a (Δt)^2
    a = -6.4m/s^2

    a= Vf - Vi / Δt -11.4

    substited -9.8 for Vf, i should be getting the same answers from each EQN shouldnt I? Unless im Rearranging wrong =/
     
  2. jcsd
  3. Mar 21, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper

    Re: AccelerationDueToGravity

    Welcome to PF.

    For no initial velocity, your displacement will be given by

    x = 1/2*g*t2

    Apparently you are dealing with experimental error. From your values calculated you have g coming in at 6.4 m/s2

    If you use g = 9.8 that implies an actual time of .694 sec. or an experimental error in measuring time of at least ± 1/6 a second. (the difference between the calculated and the actual values for g.)

    Using your a then you get v = a*t or 6.4*(.86)

    Otherwise, using your measured distance and actual g, you get v = 9.8*(.694)
     
    Last edited: Mar 21, 2009
  4. Mar 21, 2009 #3
    Re: AccelerationDueToGravity

    Wait a minute! Did the OP ever mention that the experiment was done on Earth? Perhaps, it was done on a bit lighter planet or in considerable heights above the Earth's surface.

    By Newton's second law F=m1a, and gravity force is F = (Gm1m2)/R^2 so m1a = (Gm1m2)/R^2 => a = (Gm2)/R^2 where G is the gravitational constant. So lift the object 6400 km above Earth's surface and the free fall acceleration will be around 2.45 m/s^2.
     
  5. Mar 21, 2009 #4
    Re: AccelerationDueToGravity

    Thanks for the welcome!
    I'm still a little confused,
    I've sort of convinced myself that -6.4 is my g/a, and yes it is experimental error.
    You sort of lost me after the actual time explanation tho, i dont think thats relavent for this lab tho
    I think my only question that i need answered is the correct formula and im guessing its
    x=1/2*g*t^2
    and i think i understand how you rearrange it now, you would divide x by 1/2, then divide again by t^2? which = -6.4,

    and O_O about all that gravity force~
     
  6. Mar 21, 2009 #5
    Re: AccelerationDueToGravity

    Well, gravity is a force, nothing to wonder about. What I wrote is just basics and also proof that g is independent of the object's mass that is falling towards the other object. Of course we can turn it the other way round and say that Earth is falling towards a, say, feather. However, now the Earth's mass cancels out of the equation and we get an acceleration in reference to the feather to be Tiny, with capital T because it is very small.

    About your original question, here's my approach (quite similar to yours but I'll write it using LaTex):

    S - displacement
    [itex]v_{0}[/itex] - initial velocity
    a - acceleration
    t - time

    The equation to solve your problem is as you've mentioned:

    [tex]
    S = v_{0}t + \frac{at^2}{2}
    [/tex]

    We have to solve it in terms of a, so we rearrange it:

    [tex]2S = 2v_{0}t + at^2[/tex]

    [tex]2S - 2v_{0}t = at^2[/tex]

    [tex]a = \frac{2S - 2v_{0}t}{t^2}[/tex]

    Because [itex]v_{0}[/itex] = 0 we get

    [tex]a = \frac{2S}{t^2}[/tex]

    Now input the numbers given:

    [tex]a= \frac{2 * (-2.36)}{0.86^2} \approx -6.38[/tex]

    So the object is falling with an acceleration of 6.38 [itex]\frac{m}{s^2}[/itex] which isn't much of a problem. The object needn't have an acceleration of 9.81 [itex]\frac{m}{s^2}[/itex] because of the reasons I've explained in this thread, unless it is explicitly stated that the experiment takes place close enough to the Earth's surface.

    I hope that helps.
     
  7. Mar 21, 2009 #6
    Re: AccelerationDueToGravity

    Yeah that makes a lot more sense now
    One thing I dont understtand is when you moved the 2, over to the S, why did a 2 go to the initial velocity as well? i dont remember anything liike that in algebra
    (sorry for not writing in Latex, looks complicated lol)
     
  8. Mar 22, 2009 #7
    Re: AccelerationDueToGravity

    I just multiplied both sides of the expression with 2 to get rid of the denominator. In general:

    if [tex]a = b + \frac{c}{2}[/tex], then I multiply both sides with 2 which means that I should multiply every element of the expression with 2, so

    [tex]a * 2 = b * 2 + \frac{c}{2} * 2[/tex]

    [tex]2a = 2b + c[/tex]

    That really is basic algebra.
     
  9. Mar 22, 2009 #8
    Re: AccelerationDueToGravity

    I do not recall the, "multiply every element of the expression"
    but that definetly clears up my question, and will be great to know for the rest of the year
    Thanks a lot kbaumen!
     
  10. Mar 22, 2009 #9
    Re: AccelerationDueToGravity

    No problem.

    Btw, you shouldn't try to recall sentences that a teacher has said. Rather try to understand the problem.
     
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