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Problem Involving work, velocity, given Graph

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  1. Nov 1, 2016 #1
    1. The problem statement, all variables and given/known data
    An 8kg object is moving in the positive direction of an x axis. When it passes through x=0, a constant force directed along the axis begins to act on it. The figure gives it's kinetic energy K versus position z as it moves from x=00 to x=5.0 ; K(initial)= 30 . The forces continues to act. What is v when the object moves back through x=-3m?

    2. Relevant equations


    3. The attempt at a solution
    I found the slope of the line, which is the force of the object: 30J/5m= 6N

    (0,30J) is the y intercept so y=mx+b would yield the linear eq f(x)=6N(x)+ 30J

    I plugged x=-3m into the equation to find the work at that point: f(-3)=6N(-3m)+ 30J
    W=12J

    f(0)=30 J= initial work

    W= .5m(vf)^2- 30J
    12J= .5(8kg)(vf)^2- 30J
    4J=4(vf)^2
    vf= 3.2 m/s

    My answer was fairly close to the solution manual's (3.5m/s) but I'm still unsure if my reasoning was correct. Thanks so much in advance
     

    Attached Files:

  2. jcsd
  3. Nov 1, 2016 #2

    TSny

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    Gold Member

    Welcome to PF!
    Does your slope have the correct sign?
    Note that the graph represents the kinetic energy as a function of x, not the work. So, when you substitute a value for x into the linear equation, you get the kinetic energy at that value of x.
     
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