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How do you find acceleration using Hooke's law?

  1. Sep 29, 2016 #1
    1. The problem statement, all variables and given/known data
    Imagine you were to connect your spring from this experiment to a 1.0 kg lab cart. You pull the car back and stretch the spring 30 cm, then release the car from rest, assuming there is no friction.
    A) find the acceleration at the instant when the cart is first released.
    B) When the spring's stretch has decreased to 20 cm (thus the cart has moved forward 10 cm), calculate the cart's acceleration at this instant.

    2. Relevant equations
    I have this equation that I solved for from my lab data: Force of spring = (spring constant 7.5N/M) x (the stretch of the spring s). ----> Fos (force on string) = 7.5 N/m (s)
    F=ma

    3. The attempt at a solution
    a) I know I am given the mass of the cart, initial velocity (0m/s) and the stretch of the string so I tried using f=ma and used Fos = ma.
    7.5N/m (.2 m) = (1.0kg) X a
    a = 2.25 m/s^2
    However I think it's supposed to take into consideration that the initial velocity is zero so I don't know how to apply this without the final velocity.

    B) I did force on spring = ma so 7.5(.2m) = 1.0kg x a and got 1.5 m/s^2, I want to know if I did this correctly and if I was supposed to use the fact that it moved 10cm forward.
     
  2. jcsd
  3. Sep 29, 2016 #2
    Your results look fine to me. Remember that it doesn't matter if an object starts out with 0 m/s or 1000000 m/s; any given instantaneous acceleration a = (net F)/m will only tell you how the velocity changes with that time at that instant.
     
  4. Sep 29, 2016 #3
    I am a little confused though because from my work, the units do not work out, I do not have seconds in answer and I am not given any information regarding time (seconds) so I assume I would have to incorporate initial/final velocity into my answer. Because otherwise I would be assuming m/s^2 when my work does not show that so I am not sure how to use velocity in here.
     
  5. Sep 29, 2016 #4
    I just remembered what Newtons stood for so the units cancel out, which leaves me with m/s^2 sorry. Thank you!
     
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