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Physics Question on work, energy and power

A particle in a certain conservative force field has a potential energy given by V=20xyzV = \frac{20xy}{z}. The force exerted on it is

A

(20yz)i^+(20xz)j^+(20xyz2)k^\left(\frac{20y}{z}\right)\hat{i}+\left(\frac{20x}{z}\right)\hat{j}+\left(\frac{20xy}{z^{2}}\right)\hat{k}

B

−(20yz)i^−(20xz)j^+(20xyz2)k^-\left(\frac{20y}{z}\right)\hat{i}-\left(\frac{20x}{z}\right)\hat{j}+\left(\frac{20xy}{z^{2}}\right)\hat{k}

C

−(20yz)i^−(20xz)j^−(20xyz2)k^-\left(\frac{20y}{z}\right)\hat{i}-\left(\frac{20x}{z}\right)\hat{j}-\left(\frac{20xy}{z^{2}}\right)\hat{k}

D

(20yz)i^+(20xz)j^−(20xyz2)k^\left(\frac{20y}{z}\right)\hat{i}+\left(\frac{20x}{z}\right)\hat{j}-\left(\frac{20xy}{z^{2}}\right)\hat{k}

Answer

−(20yz)i^−(20xz)j^+(20xyz2)k^-\left(\frac{20y}{z}\right)\hat{i}-\left(\frac{20x}{z}\right)\hat{j}+\left(\frac{20xy}{z^{2}}\right)\hat{k}

Explanation

Solution

Given : V=20xyzV=\frac{20 x y}{z}
For a conservative field
F⃗=−∇⃗V;\vec{F}=-\vec{\nabla} V ; where, ∇⃗=i^∂∂x+j^∂∂y+k^∂∂z\vec{\nabla}=\hat{i} \frac{\partial}{\partial x}+\hat{j} \frac{\partial}{\partial y}+\hat{k} \frac{\partial}{\partial z}
∴F⃗=[i^∂V∂x+j^∂V∂y+k^∂V∂z]\therefore \vec{F}=\left[\hat{i} \frac{\partial V}{\partial x}+\hat{j} \frac{\partial V}{\partial y}+\hat{k} \frac{\partial V}{\partial z}\right]
=−[i^∂∂x(20xyz)+j^∂∂y(20xyz)+k^∂∂z(20xyz)]=-\left[\hat{i} \frac{\partial}{\partial x}\left(\frac{20 x y}{z}\right)+\hat{j} \frac{\partial}{\partial y}\left(\frac{20 x y}{z}\right)+\hat{k} \frac{\partial}{\partial z}\left(\frac{20 x y}{z}\right)\right]
=−(20yz)i^−(20xz)j^+(20xyz2)k^=-\left(\frac{20 y}{z}\right) \hat{i}-\left(\frac{20 x}{z}\right) \hat{j}+\left(\frac{20 x y}{z^{2}}\right) \hat{k}