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Quick clarification on equation of motion

  1. Mar 26, 2013 #1
    A 900N force acts on a 5kg box in horizontal direction initially at rest. When V=6m/s the force stops.
    Through what distance was the block pushed.

    I said this:
    F=ma
    v=x/t ---> t=x/v (1)
    a=v/t (2)
    combine (1) and (2)
    a=v^2/x
    So x=v^2/a

    That is the same as angular circular motion.

    Then there is
    x=0.5at^2
    With t=x/v
    x=0.5a(x/v)^2
    1=0.5ax/v^2
    2=ax/v^2
    x=(2v^2)/a

    Where am I screwing up?

    Thank you for your time.
     
  2. jcsd
  3. Mar 26, 2013 #2
    v = x/t is valid only for non-accelerated motion, which is not the case here.

    a = v/t is correct.
     
  4. Mar 26, 2013 #3
    Thank you for your response Voko :)

    I'm an idiot.
    This is why I feel hesitant to help people on short notice (after being woken up no less!) because I get all overwhelmed and want to make sure everything is perfect and my mind starts making stupid mistakes.



    So the equation x=(2v^2)/a needs to be reworked because I used v=x/t again.
     
    Last edited: Mar 26, 2013
  5. Mar 26, 2013 #4
    so x=v^2/(2a)..

    x=(at^2)/2

    so t=v/a
     
  6. Mar 26, 2013 #5
    So for going up an incline with friction,

    It has an initial known KE, thus v

    a comes from Fnet=ma

    Fnet=u(mg)cos(60)+(mg)cos(60)

    Then x=v^2/(2a)
    So x is the displacement up the incline correct?

    Why is mind getting on the fact that velocity is positive and a is negative? but displacement shouldn't be negative
     
  7. Mar 26, 2013 #6
    The formula you derived in #4 is valid for motion from rest till some given velocity. In #5, however, it is the other way around.
     
  8. Mar 26, 2013 #7
    I'm sorry I don't know what you mean by other way around.


    But for #5 up the incline, wouldn't it be x=[(vf2)-(vi2)]/(2a)

    So negative v on top, negative a on bottom. That takes care of the negative.

    But I still don't understand if this can be used to find x up an incline. What is the other way around?
     
  9. Mar 26, 2013 #8
    The other way around is you start with non-zero velocity, and end up with no velocity. The new formula you have is the correct one, it gives you the correct sign of the displacement.
     
  10. Mar 26, 2013 #9
    Ok I see what you're saying now..

    Thank you very much!
     
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