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Acceleration of a rock

  1. Jan 29, 2010 #1
    I used to know this stuff. How fast will a rock be going after falling 5 seconds in a vacuum?
    I'd also like to know the formula in simple terms.

    At the gym on the hip sled we call it- you lay back and push up with feet and legs. The rack supporting the weights is at 45 degrees. If you have 500# on it how much force is required to move it? I'd also like the formula for this as well as the answer. I think it's a simple 50% or 250# because it's at 45 degrees but not sure. I just don't like these kids bragging to an old guy that they are pushing 500#!

    hip sled here:


    Thank you! Tregg
  2. jcsd
  3. Jan 29, 2010 #2
    if its 45 degrees. what is sin(45)? its more than 50%. Anyways, the answer is Fgsin(45).
  4. Jan 29, 2010 #3
    If the mass of the "weight" is M, gravitational acceleration is g, and the angle between the inclined plane and the ground is A, then the force necessary to lift the weight is

    F = M*g*sin(A)

    This corresponds to lifting a mass M' = F/g = M*sin(A)

    M=100kg, g=10m/s^2, A=45 degrees gives M' = 100*sin(45) kg

    Now, sin(45) = sqrt(2)/2, which is approx 0.7, so M' = 70 kg in this case.

    If the angle A was 30 degrees, it would be half, since sin(30) = 1/2.

  5. Jan 29, 2010 #4


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    As for your question on the falling rock, I'm going to assume you mean on the surface of Earth, where gravity accelerates the rock at 9.8 m/s^2

    thus for 5 seconds we have:

    [tex] v = at = (9.8 \frac{m}{s^2}) (5 s) = 49 \frac{m}{s} = 110 MPH [/tex]
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