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:

Question

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???

Answer 1

INT Fdx= U

F= dU/dx= you do it.

Answer 2

Hannah. Isnt costheta=1 here?

U=-INTF.dx

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

Answer 3

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

Answer 4

YEAH that's it, thanks bobpursley

Answer 5

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

Answer 6

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)

Answer 7

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:

INT Fdx= U

Hannah. Isnt costheta=1 here?

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

YEAH that's it, thanks bobpursley

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

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)

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