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Finding force constant, not given the value of k

  1. Nov 16, 2011 #1
    A 0.650 kg block slides on a frictionless, horizontal surface with a speed of 1.19 m/s. The block encounters an unstretched spring and compresses it 25 cm before coming to rest.
    (a) What is the force constant of this spring? ____ N/m

    (b) For what length of time is the block in contact with the spring before it comes to rest? ___s

    For part a:
    Since the formula for a force constant is f=-kx, and the problem doesn't give a k value, I tried solving for it using:

    k=(mg)/x and got
    k=(.650kg*9.81)/.25 m
    k=25.50g

    and plugged this in to f=-fx
    and got f=-6.3765

    For part b:
    im not even sure where to start ..
     
  2. jcsd
  3. Nov 16, 2011 #2
    I do not think that is correct. The force compressing the spring is NOT the weight of the mass mg.

    Try thinking about conservation of energy.
    What is the formula for elastic PE in a spring?
     
  4. Nov 16, 2011 #3
    I got the answer for part a:

    i used
    v^2=u^2+2as
    and just continued on from there..

    how do i start doing part b?
     
  5. Nov 16, 2011 #4
    Note that in using 'v[itex]^{2}[/itex]=u[itex]^{2}[/itex]+2as' one has to use the AVERAGE acceleration since in this case the acceleration is varying.
     
    Last edited: Nov 16, 2011
  6. Nov 16, 2011 #5
    I used v^2=u^2as
    then resultant force = k X extension
    .650*(-2.8322)-0=k(.25)
    k=14.7

    (i was originally reviewing this problem and also given this answer, but I was having a hard time getting to the answer/didn't know how to start, the answer given for part a was 14.7)
     
  7. Nov 16, 2011 #6
    For (b) one can use 'v = u + at' . But be careful to use the AVERAGE acceleration since
    'v = u + at' can only be used when acceleration is constant or when the average acc is being used.
     
  8. Nov 16, 2011 #7
    average acceleration is v/change in time, but i dont have the change in time? do i use the change in distance: .25m?
     
  9. Nov 16, 2011 #8
    You know v and you also know a=2.8322m/(s^2) and so you can find t. But use the average acceleration which in our case is a/2 since the force is directly proportional to the extension.
     
  10. Nov 16, 2011 #9
    this is what im doing:

    v=u + at
    1.4161 = 1.19 + 1.4161t
    .2261=1.4161t
    t=.15966

    but the correct answer is said to be t=.33 s
     
  11. Nov 16, 2011 #10
    I keep getting t = 0.42s.
     
  12. Nov 16, 2011 #11
    note that v = 0 and average acc is 2.8269m/s^2.
     
  13. Nov 17, 2011 #12
    Now I notice what I did wrong.
    The acceleration is not constant since the displacement is one quarter of an shm. So the time is one quarter of the period and the period is 2piSquareRoot(m/k).
     
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