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Velocity, Work, and Acceleration

  1. Sep 18, 2010 #1
    For number 2, I cant seem to figure it out. Im using the equation from deltaKinetic Energy which is v=sq(2F)(dx))

    The F for force I'm using is the spring constant x displacement which is 4886(.447). dx=.447. Take the sqroot of that and you get 44.18, but thats incorrect. I tried so many values that I lost track. Not really sure how I should go about this one?
  2. jcsd
  3. Sep 18, 2010 #2
    Hi mohabitar

    Try using work = change in kinetic energy.
  4. Sep 18, 2010 #3
    Ok thanks got this one (7.21 was the answer).
    Now, same problem, different question:

    This one has stumped me. I tried using the same process as I did for the other one, but Im just not really sure how to factor fraction in there. Am I supposed to find the acceleration of the box during the non-friction area and subtract from it the acceleration during friction to find net acceleration? Ah I honestly dont even know where I can start this one.
  5. Sep 18, 2010 #4
    No, you got the wrong idea. The block is not accelerating when it has been released from the spring. Instead, it moves with constant speed because there is net force acting on it. The spring force only acting on it when it is attached to it and when the spring is compressed.

    So you only need to find the acceleration caused by friction (which is called deceleration). Using Newton's second law, then use kinematics to find the speed
  6. Sep 18, 2010 #5
    Ok so F[spring]=SpringConstant(Normal Force)=.4(mg)=.4(16)(9.81)=62.78
    Then I used the kinematics equation V^2=u^2+2as putting s as 2 and used the above value for a, but that wasnt working out. Am I on the wrong path?
  7. Sep 19, 2010 #6
    That's not the right equation (although maybe you got the same result as the right equation). Using Newton's second law:
    ΣF = ma

    What is the resultant force acting on the object (in horizontal direction) when it slides through the rough floor?
    Don't forget the value of a should be negative when you plug it to kinematic formula.
  8. Sep 19, 2010 #7
    Still not getting this. What do I use newtons second law to find? A by friction? So F=ma, I need to find a and I dont know F, so how would I solve for a? And what what delta x be in this situation, just 2? When I use a kinematics equation, will I have an initial speed of 7.21?
  9. Sep 19, 2010 #8
    Hi mohabitar

    Yes, you use Newton's second law to find the deceleration because you need it to find the final speed after getting through the rough floor. The force acting on the object (in horizontal direction) is friction only.

    I am not sure what you mean by Δx, but if you mean it is the distance to put in the kinematics formula, yes it is 2 m; and the initial speed is indeed 7.21
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