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Finding speed from force-vs-position graph

  1. Mar 17, 2012 #1
    1. The problem statement, all variables and given/known data

    The force shown in the figure (url below) acts on a 1.7-kg object whose initial speed is 0.44 m/s and initial position is x=0.27 m.

    http://session.masteringphysics.com/problemAsset/1123629/2/Walker4e.ch07.Pr040.jpg

    Part A: Find the speed of the object when it is at the location x=0.99m.

    Answer: 0.76 m/s

    Part B: At what location would the object's speed be 0.32 m/s?

    2. Relevant equations

    Vf^2=Vi^2 + 2aΔx

    W=F(Δx)


    3. The attempt at a solution

    I already answered Part A correctly, using the position/velocity kinematic equation along with the Work=Force x distance equation by using the given mass and the average of the graphed forces through the given distance. For that acceleration, I had 0.275 m/s^2. For Part B, I tried two things, first I used the same kinematic equation with the same acceleration only changed the final velocity to 0.32 m/s. I added the delta x to the 0.27m in the problem and got the wrong answer. Then, I tried setting the initial velocity as that of part A and solving it that way but I still go the wrong answer. I think Im a little confused as to whether the object is slowing down or speeding up and whether or not the problem began after an initial push from rest or is taking place after the object has already been moving. Someone please help me. Thanks.
     
  2. jcsd
  3. Mar 17, 2012 #2

    PeterO

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    Homework Helper

    Re: Can anyone help me with this problem?

    Th initial position/velocity/acceleration presumably refers to what was happening when we chose to look at the object. [After all, if we wanted to analyse the motion of cars on a freeway, it doesn't mean none of them are allowed to move until we look at them].

    The speed of 0.32, being smaller that the "initial" speed is 0.44 presumably means the answer sought is a negative time, when the object was closer to the origin ; perhaps at 0.1m or a tiny fraction less.
     
  4. Mar 17, 2012 #3
    Figured it out. Thanks for your help :)
     
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