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Finding velocity from force and distance

  1. Oct 21, 2012 #1
    1. The problem statement, all variables and given/known data
    mass=60kg
    distance=9.6m

    Find velocity

    2. Relevant equations
    force=mass*acceleration

    3. The attempt at a solution

    I multiplied mass by accelleration(9.8) and got 588. Thats the force.

    What do I do with force and distance to get velocity?
     
  2. jcsd
  3. Oct 21, 2012 #2
    Could you be more specific? Are you dropping a mass from a given height and trying to find the velocity upon impact?

    The most important part of physics is learning how to state a problem. This may be your issue.
     
  4. Oct 21, 2012 #3
    Yes. I'm trying to find the velocity on impact.

    Sorry for the inconvience
     
  5. Oct 21, 2012 #4
    Do you have to use force? I'd use the principle of conservation of energy for this problem. It practically screams for it.
     
  6. Oct 21, 2012 #5
    What is that?

    I haven't run into that yet.
     
  7. Oct 21, 2012 #6
    Oh.

    Hmm, well, this is the advanced physics section, so I'd assumed you had.

    Are you in calc-based physics? If so, you might use f=ma and use the fact that a=dv/dt=(dv/ds)(ds/dt), and recognize that ds/dt is v. This should yield an expression for v using only distance as your independent parameter.
     
  8. Oct 21, 2012 #7
    Yes I'm in calc-based physics.

    Because it's being dropped from zero velocity wouldn't the acceleration be gravity(9.8)?

    If so, are you saying that I should solve for ds/dt?
     
  9. Oct 21, 2012 #8
    No.

    If a=9.8, then you need to integrate the expression dv*v=a*ds to find an expression for v, where ds is a differential arc length (luckily in your case you can assume that the arc has no curvature, and so your line integral breaks down into just an ordinary integral as you're familiar with - in other words, s is just the height). You are just using the chain rule as described in post #6 to create an expression that uses the variables you know, i.e. distance and acceleration, to find one you don't, i.e. velocity.

    The resulting equation should be in your book, but now you know how it's derived.
     
  10. Oct 21, 2012 #9
    Thank you so much I understand now. But one remaining question.

    in"dv", does that mean distance*velocity or the velocity traveled in the distance?
     
  11. Oct 21, 2012 #10
    That's actually a differential unit of velocity. Like in calculus the expression "dx" is a differential length in the x direction. Same thing here. Since velocity is always changing, you want to isolate an infinitesimally small unit of velocity such that the velocity is "constant" over that infinitesimally small period of time (or in this case, distance, since the differential is dv/ds). This allows you to sum up all the infinitesimally small "chunks" of velocity - which is what an integral does.

    The extra v term is used just like you would if you were integrating x*dx.

    Sorry if this is confusing in any way. I realize now that maybe this derivation may be a little much. But it's EXTREMELY HANDY to know how these equations of motion are derived. I didn't really learn kinematics until I knew the derivations.
     
  12. Oct 21, 2012 #11
    ok thanks
     
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