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Weigh Problem

  1. Oct 1, 2008 #1
    1. The problem statement, all variables and given/known data
    When venturing forth on Planet X, you throw a 5.24kg rock upward at 13.0 m/s and find that it returns to the same level 1.51 s later.

    What does the rock weigh on Planet X?
    2. Relevant equations

    W=m*a

    3. The attempt at a solution

    To find the acceleration i good the 13.0m/s / 1.51s = 8.61 m/s^2. W=5.24kg * 8.61 m/s^2 = 45.1 N

    The online hw keeps telling me that my answer is wrong. What am I doing wrong? Thank you for the help!
     
  2. jcsd
  3. Oct 1, 2008 #2
    Try using this equation:
    x = (x-initial) + (V-initial)(t) + (1/2)(a)(t^2)
    (remember that in your case the change in x = 0, which means x - (x-initial) = 0)

    Someone correct me if I am wrong :0
     
  4. Oct 1, 2008 #3
    With this formula could be easier: Vf=Vo+at.
    Vo is your initial velocity, Vf is the final velocity that for you is the velocity at highest point the stone reaches (which is 0), and the time for you is 1.52/2 because you're just dealing with half of the experiment.
     
  5. Oct 1, 2008 #4

    Mentallic

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    Homework Helper

    You can use: [tex]s=ut+\frac{1}{2}at^{2}[/tex] for the entire 'experiment'.

    [tex]s=displacement[/tex]
    [tex]u=initial (starting) velocity[/tex]
    [tex]t=time[/tex]
    [tex]a=acceleration[/tex]

    Note: You are trying to find the acceleration so the weight formula can be used later.

    Therefore, by rearrangement, [tex]a=\frac{2(s-ut)}{t^{2}}[/tex]

    Since the rock is launched at 13ms-1, this is the initial velocity (u)
    The time taken for the rock to come back from where it was launched, t=1.51 seconds
    This seems to be all the information given that is relevant to this equation. But it is also known that the displacement (s) will be 0 when the rock reaches back to where it left.

    Hence, [tex]a=\frac{2(0-13(1.51))}{(1.51)^{2}}[/tex]

    [tex]a \approx -17.22 ms^{-2}[/tex] (note: negative acceleration means the acceleration is acting opposite to the rock launch. i.e. downwards towards the planet)

    Now since the acceleration has been found, the weight of the rock can be found using:
    [tex]F=ma[/tex]

    F= force (newtons) / weight of object
    m = mass = 5.24kg
    a = acceleration = -17.22 ms-2
    Therefore, [tex]F=(5.24)(17.22)[/tex]

    [tex]F\approx 90kg[/tex]
     
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