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Potential Energy vs. Position Graphs

  1. Feb 18, 2017 #1
    1. The problem statement, all variables and given/known data
    A particle has the potential energy shown in the figure. (Figure 1) What is the
    x-component of the force on the particle at x =5, 15, 25, and 35 cm?

    I'm stuck on finding the Fx at 25 & 35 cm.


    2. Relevant equations

    F = -du/dx
    100 N = 1J/cm



    3. The attempt at a solution

    Slope throughout 20-40 cm is a constant positive .5.

    So at 25 cm it should be :
    F = -.5/25cm * 100 = -2 N

    This is incorrect. I'm really confused?
     

    Attached Files:

  2. jcsd
  3. Feb 18, 2017 #2
    Hey! You're spot on that the slope is constant throughout the 20cm-40cm range which means the average rate of change in that region is equal to the instantaneous rate of change at all points in the region. You also correctly pointed out that the expression for force is the negative or the rate of change of energy. So if the rate of change (slope) is constant in that region, what does that imply about the force at all points in the region? If you think about that question it should reveal that your problem lies in dividing by 25 cm. It's also a good idea to check your units if you run into these problems. Dividing the slope by a distance unit doesn't give you a Newton does it?
     
  4. Feb 18, 2017 #3
    Hello,
    so i calculated the force applied at 25cm to be -2 N which is incorrect :( I'm just really confused as to what is wrong as i feel like i derived it correctly
     
  5. Feb 18, 2017 #4
    Let's go back to the basic equation. So, F=-dU/dx. If the slope is constant then this can be written as F=-[U(40)-U(20)]/[40-20] (then multiply by 100 to convert from cm to m) You could use any points in the interval, but will get the same value because, as you pointed out, the slope is constant (the same over whole interval)
     
  6. Feb 18, 2017 #5
    Ohh okay thank you hahah i get it now. The force will be constant everywhere that makes more sense now
     
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