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

Question

The potential energy (in joules) function of a particle in a region of space is given as

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

Here $x, y, z$ are in meters. Find the magnitude of the x-component of force (in newton)

acting on the particle at point $P (1 m, 2 m, 3 m)$ .

TestHub · Accepted Answer

The potential energy is given by U = 2 x 2 + 3 y 3 + 2 z . The force component in the x-direction is given by F x ​ = − ∂ x ∂ U ​ . Taking the partial derivative of U with respect to x , we get F x ​ = − 4 x . At the point P ( 1 , 2 , 3 ) , x = 1 , so F x ​ = − 4 ( 1 ) = − 4 N. The magnitude of the x-component of the force is ∣ F x ​ ∣ = ∣ − 4∣ = 4 N.

Physics - W.E.P. Question with Solution | TestHub

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