Potential Energy question

2021-10-11

When a particle is a distance r from the origin, its potential energy function is given by the equation U(r)=kr, where k is a constant and \(r=x^2+y^2+z^2\)

(a) What are the SI units of k?

(b) Find a mathematical expression in terms of x, y, and z for the y component of the force on the particle.  

(c) If U=3.00 J when the particle is 2.00 m from the origin, find the numerical value of the y component of the force on this particle when it is at the point (-1.00 m, 2.00 m, 3.00 m).

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Expert Answer

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Answered 2021-10-14 Author has 1944 answers

First convert to cartesian
\(U(x,y,z)=k_r=k(x^2+y^2+z^2)^{1/2}\)
Now, in vague terms, the force is sort of, the "rate of change" of the potential energy function. E.g. gravity. The stronger the force, the more work you have to do to move upwards against it, and the more potential energy you gain as a result. So, from this, we can skip a bunch of math and take the leap to saying that the force in the Y direction depends only upon how U changes *in the y direction*. "How U changes in the y direction" is basically what the partial derivative of U with respect to y tells you. In general, I need four dimensions to plot U(x,y,z). However, in that 4D space, if I choose to move only in the y direction (*keeping x and z constant*), the rate of change of U that I measure *along that line* is the partial derivative of U with respect to y. Mathematically:
\(F_y=U(x,y,z)y=y_k(x^2+y^2+z^2)^{1/2}\)

= the derivative of U(x,y,z) with respect to y, *assuming that x and z are constants*
This is not the total rate of change of U, but only the rate of change you'd measure if you confined yourself to movement in the y direction. Hence, it's a partial derivative.

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