1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Acceleration through slope

  1. Jul 11, 2007 #1
    1. The problem statement, all variables and given/known data
    If the slope of a graph of the acceleration of a cart vs. the number of paper clips attached by a string and suspended over a pulley is ( 20 cm/s2) / clip, and if the slope of a graph of number of paper clips needed to maintain equilibrium vs. ramp slope is 47 clips / unit of ramp slope, then how many cm/s2 of acceleration should correspond to 1 unit of ramp slope? If 52 clips are necessary to match the mass of the cart, then if we could apply this force to the cart without the extra mass of all those clips, what would be the acceleration of the cart?

    2. Relevant equations
    I was thinking of using maybe one of the constant acceleration equations.
    v = v0 + a t
    x = x0 + v 0 t + 1/2 a t ^ 2
    v ^ 2 = v 0 ^ 2 + 2 a ( x - x 0 )
    avg v = ( v+ v 0 ) / 2

    v = velocity
    v 0 = initial velocity
    a = acceleration
    t = time ( seconds )
    x0 = initial position
    x = position

    3. The attempt at a solution
    I am completely confused with this question. I have put in over 10 hours, and tried to find out help all over the internet. I am looking for alot more than just the answer, because I want to learn this stuff. If someone could walk me through this I would greatly appreciate it.
  2. jcsd
  3. Jul 12, 2007 #2
    Here is something I worked out earlier, can I get some feedback.

    If you get 18cm/s^2 of acceleration per clip and you get equilibrium per unit of slope for 59 clips it seems reasonable that a unit of slope produces:

    18 * 59 cm/s^2 = 1062 cm/s^2

    The cart has a mass of 57 clips so the difference is 2 clips? 2 * 18 = 36 cm/s^2 ?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook