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A 10-kilogram body is constrained to move along the x-axis. The potential energy U of the body in joules is given as a function of its position x in meters by:
U(x) = 6x^2 - 4x + 3
The force on the particle at x=3 meters is?

Here was my thinking.... U=-W. So, W(x) = -6x^2 + 4x - 3. W also = Fxcos(theta). So I plugged in x=3 to all x's to get Fcos(theta) = -3, so the force would be 3N to the left. Why is this incorrect????? Am I thinking way off track???

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Hannah

  1. Yeah you're a little off track but not by far someone a little more knowledgeable would be able to give the correct formula and answer

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    bill j.

  2. I just realized I totally forgot to include the mass. Can anyone help???

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    Hannah

  4. INT Fdx= U
    F= dU/dx= you do it.

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    bobpursley

  5. YEAH that's it, thanks bobpursley

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    bill j.

  6. I thought U was the negative integral of Fdxcos(theta)? But I guess I was wrong...

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    Hannah

  8. OK now it makes sense. The cos(theta) doesn't come in until the end, right? I got an F and then it's either neg or pos according to the angle. Thanks Bob (and Bill)

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    Hannah

  9. Hannah. Isnt costheta=1 here?
    U=-INTF.dx

    The negative sign takes care of direction of F (it is a vector). Think about that.

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    bobpursley

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