The potential function associated with a force P in space is known to be V(x, y, z) = -(x²+y²+z²)^1/2. (a) Determine the x, y, and z components of P. (b) Calculate the work done by P from O to D by integrating along the path OABD, and show that it is equal to the negative of the change in potential

Question

The potential function associated with a force P in space is known to be [latex]V\left( x,y,z \right) =-{ \left( { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } \right) }^{ 1/2 }[/latex] (a) Determine the x, y, and z components of P. (b) Calculate the work done by P from O to D by integrating along the path OABD, and show that it is equal to the negative of the change in potential from O to D.

Accepted Answer

(a) [latex]P_{x}=-\frac{\partial V}{\partial x}=-\frac{\partial\left[-\left(x^{2}+y^{2}+z^{2} ight)^{1 / 2} ight]}{\partial x}=x\left(x^{2}+y^{2}+z^{2} ight)^{-1 / 2}\blacktriangleleft[/latex]

[latex]P_{y}=-\frac{\partial V}{\partial y}=-\frac{\partial\left[-\left(x^{2}+y^{2}+z^{2} ight)^{1 / 2} ight]}{\partial y}=y\left(x^{2}+y^{2}+z^{2} ight)^{-1 / 2}\blacktriangleleft[/latex]

[latex]P_{z}=-\frac{\partial V}{\partial z}=-\frac{\partial\left[-\left(x^{2}+y^{2}+z^{2} ight)^{1 / 2} ight]}{\partial z}=z\left(x^{2}+y^{2}+z^{2} ight)^{-1 / 2}\blacktriangleleft[/latex]

(b) [latex]U_{O A B D}=U_{O A}+U_{A B}+U_{B D}[/latex]

[latex]O-A: P_{y}[/latex] and [latex]P_{x}[/latex] are perpendicular to O - A and do no work.

Also, on O - A [latex]\quad x=y=0 \quad ext { and } \quad P_{z}=1[/latex]

Thus, [latex]\quad U_{O-A}=\int_{0}^{a} P_{z} d z=\int_{0}^{a} d z=a[/latex]

[latex]A-B: P_{z}[/latex] and [latex]P_{y}[/latex] are perpendicular to A - B and do no work.

Also, on A - B [latex]\quad y=0, \quad z=a \quad ext { and } \quad P_{x}=\frac{x}{\left(x^{2}+a^{2} ight)^{1 / 2}}[/latex]

Thus,

[latex]\begin{aligned}U_{A-B} & =\int_{0}^{a} \frac{x d x}{\left(x^{2}+a^{2} ight)^{1 / 2}} \& =a(\sqrt{2}-1)\end{aligned}[/latex]

[latex]B-D: P_{x}[/latex] and [latex]P_{z}[/latex] are perpendicular to B - D and do no work.

On B - D,

[latex]\begin{aligned}k & =a \z & =a \P_{y} & =\frac{y}{\left(y^{2}+2 a^{2} ight)^{1 / 2}}\end{aligned}[/latex]

Thus,

[latex]\begin{aligned}U_{B D} & =\int_{0}^{a} \frac{y}{\left(y^{2}+2 a^{2} ight)^{1 / 2}} d y=\left.\left(y^{2}+2 a^{2} ight)^{1 / 2} ight|_{0} ^{a} \U_{B D} & =\left(a^{2}+2 a^{2} ight)^{1 / 2}-\left(2 a^{2} ight)^{1 / 2}=a(\sqrt{3}-\sqrt{2}) \U_{O A B D} & =U_{O-A}+U_{A-B}+U_{B-D} \& =a+a(\sqrt{2}-1)+a(\sqrt{3}-\sqrt{2})\quad\quad U_{O A B D}=a \sqrt{3}\blacktriangleleft \\Delta V_{O D} & =V(a, a, a)-V(0,0,0) \& =-\left(a^{2}+a^{2}+a^{2} ight)^{1 / 2}-0 \quad\quad \Delta V_{O D}=-a\sqrt{3}\blacktriangleleft\end{aligned}[/latex]

Thus,[latex]\quad U_{O A B D}=-\Delta V_{O D}[/latex]

Question 13.82: The potential function associated with a force P in space is...

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