A particle in a conservative force field has a potential energy given by U=(20xyz). The force exerted on it is

Given, U=20xyz−1 Fx=−∂u∂x=−20yz Fy=−∂u∂y=−20xz and Fz=−∂u∂z=20xyz2 ∴F→=Fxi^+Fyj^+Fzk^ =−(20yz)i^−(20xz)j^+(20xyz2)k^

A particle in a conservative force field has a potential energy given by U=(20xyz). The force exerted on it is

Given, U=20xyz−1 Fx=−∂u∂x=−20yz Fy=−∂u∂y=−20xz and Fz=−∂u∂z=20xyz2 ∴F→=Fxi^+Fyj^+Fzk^ =−(20yz)i^−(20xz)j^+(20xyz2)k^

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A particle in a conservative force field has a potential energy given by $U = (\frac{20 x y}{z})$ . The force exerted on it is

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$(\frac{20 y}{z}) \hat{i} + (\frac{20 x}{z}) \hat{j} + (\frac{20 x y}{z^{2}}) \hat{k}$

$- (\frac{20 y}{z}) \hat{i} - (\frac{20 x}{z}) \hat{j} + (\frac{20 x y}{z^{2}}) \hat{k}$

$- (\frac{20 y}{z}) \hat{i} - (\frac{20 x}{z}) \hat{j} - (\frac{20 x y}{z^{2}}) \hat{k}$

$(\frac{20 y}{z}) \hat{i} + (\frac{20 x}{z}) \hat{j} - (\frac{20 x y}{z^{2}}) \hat{k}$

answer is B.

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Detailed Solution

Given, $U = 20 x y z^{- 1}$
$F_{x} = - \frac{\partial u}{\partial x} = - \frac{20 y}{z} F_{y} = - \frac{\partial u}{\partial y} = - \frac{20 x}{z}$
and $F_{z} = - \frac{\partial u}{\partial z} = \frac{20 x y}{z^{2}}$
$∴ \vec{F} = F_{x} \hat{i} + F_{y} \hat{j} + F_{z} \hat{k} = - (\frac{20 y}{z}) \hat{i} - (\frac{20 x}{z}) \hat{j} + (\frac{20 x y}{z^{2}}) \hat{k}$