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Object launching off table

  1. Feb 26, 2008 #1
    1. The problem statement, all variables and given/known data

    A popular pastime is to see who can push an object closest to the edge of a table without its going off. You push the 100 g object and release it 2.20 m from the table edge. Unfortunately, you push a little too hard. The object slides across, sails off the edge, falls 0.500 m to the floor, and lands 25.0 cm from the edge of the table.

    If the coefficient of kinetic friction is 0.300, what was the object's speed as you released it?
    2. Relevant equations


    y = (1/2)*g*t^2 therefore t = (2*y/g)^(1/2)
    x = v2*t therefore x = v2*(2*y/g)^(1/2)

    3. The attempt at a solution

    I worked everything out and I got 3.60 m/s for the speed when released but my online hw program says "Not quite. Check through your calculations; you may have made a rounding error or used the wrong number of significant figures." I checked my work and Im not sure where I went wrong.

    Here are the values I got

    V2=.7836 m/s velocity when it leaves the table
    T(3-2) time it takes to hit the ground once leaving the table=.319 sec
    A0=-2.94m/s^2
    V0=3.5966

    To find V0 I used X1=(-Vox^2)/(2Ax) then solved for Vox any insight as to where I went wrong?
     
  2. jcsd
  3. Feb 26, 2008 #2
    I thought I'd give this problem a shot, so here it goes. First, I find the time it takes for the object to hit the ground once its off the edge of the table:

    0.500 m = (1/2) (9.8 m/s^2) t^2 -> t = Sqrt(1/9.81) -> t = 3.19E-1 s

    next, I solve for the velocity in the x direction as the object leaves the table:

    .25 m = Vedge (3.19E-1 s) -> Vedge = .25/3.19E-1 -> Vedge = 7.83E-1 m/s

    Here is where it gets fuzy, I solved for Vinitial in terms of energy:

    (1/2) m Vinitial^2 - F d = 1/2 m Vedge^2 -> Vinitial = Sqrt(Vedge^2 +2 Fd/m)
    with Fd/m = (mue) g d = .3*9.8*2.20

    So Vinitial = Sqrt((7.83E-1)^2 + 2(6.468)) = 3.68 m/s

    Assuming I'm right, your answer is close. You probably rounded off at some point. I think significant figures are only 2, due to .25 being given, so the answer would round off to 3.7? Significant figures are not my strong point. Take this with a grain of salt.
     
  4. Feb 26, 2008 #3

    Dick

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    Yes. You should be using v0^2-v2^2=2*a*x.
     
  5. Feb 26, 2008 #4

    Dick

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    bhimberg is correct. You should be taking the difference of the velocities squared equal to 2*a*x. It's not round off, it's a concept problem.
     
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