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The potential energy function of a particle in a region of space is given as:
$U = (2 x^{2} + 3 y^{3} + 2 z) J$
Here x, y and Z are in meters. Find the force acting on the particle at point P(1 m, 2 m, 3 m):

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$- (4 \hat{i} + 36 \hat{j} + 2 \hat{k}) N$

$- 8 \hat{i} - 18 \hat{j} + 2 \hat{k}$

$- 4 \hat{i} + 18 \hat{j} - 9 \hat{k}$

$- 4 \hat{i} + 36 \hat{j} + 9 \hat{k}$

answer is A.

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Detailed Solution

$U_{2} = (2 x^{2} + 3 y^{3} + 2 z)$

$\Rightarrow \vec{F} = - (\frac{\partial U}{\partial x} \hat{i} + \frac{\partial U}{\partial y} \hat{j} + \frac{\partial U}{\partial z} \hat{k})$

$= - (4 xi + 9 y^{2} \hat{j} + 2 \hat{k}) N$

$so, {\vec{F}}_{(1, 2, 3)} = - (4 \hat{i} + 36 \hat{j} + 2 \hat{k}) N$

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