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Work & KE Problem

  1. Feb 19, 2009 #1
  2. jcsd
  3. Feb 19, 2009 #2


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    Well, according to the rules you're supposed to show your efforts to solve this. Since you didn't do that, I assume you don't have any to show. I'll get you started:

    (1) We'll assume friction is being ignored. Then the work done goes into generating kinetic energy.

    (2) Work done is given by

    [tex] W = \int \mathbf{F} \cdot d \mathbf{l} [/tex]

    (3) Since the force is not constant you need the equation for the force.

    That should get you started.
  4. Feb 19, 2009 #3
    oops i forgot that xD

    I tried to solve this using a work energy theorem

    SO because it is starting at rest the KE initial is zero

    THerefore the equation would be W =.5mv^2

    When solving for v, I get like 10.59, not an answer near the possible ones
  5. Feb 19, 2009 #4
    Also we havent learned integrals or whatever that L is in the equation you gave me, is there an easier way to start off?
  6. Feb 19, 2009 #5


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    Well, there is. You can approximate the integral in the following way: Break the horizontal distance from 0 to 150 up into convenient length segments, for starters, say 15 meters wide. That will give you 10 segments. From the graph estimate the average force over that segment (I'd pick the midpoint of the segment). Multiply that value by 15 meters. That gives you a
    [tex] \Delta W[/tex] for that segment. Do this 10 times, once for each segment and add them up. That's an approximation to the total work. Set that equal to the kinetic energy and solve for v.

    Three comments:

    (1) The more segments you take the better the approximation.

    (2) I just thought of this --you can take the midpoint of your segment and substitute it into the equation of the line to get an approximate average force for that segment

    (3) What you are doing here is a simple graphical integration. Keep it in mind when you study calculus
  7. Feb 19, 2009 #6
    Could I just find the area under the line?
  8. Feb 19, 2009 #7
    Awesome, I used an intergral function on my calculator to find its total work, then set that to v. Thanks a ton!
  9. Feb 19, 2009 #8


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    Well, duh. Why didn't I think of that?

    Answer --I was too wrapped up in the line integral idea.
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