Question

The potential energy function for a two dimensional
force is U=+2x3 - 4x²y +6y2 where U is in joules, x and
y are in meters. The force (in N) acting on particle at (3,2)
is Ai + Bj. Find A+B.​

Answer 1

Given:

The potential energy function for a two dimensional force is U=+2x³ - 4x²y +6y². The force (in N) acting on particle at (3,2) is Ai + Bj.

To find:

Value of A + B.

Calculation:

We will perform partial derivation of the potential energy function to get the the force function.

For x axis:

$U = 2 {x}^{3} - 4 {x}^{2} y + 6 {y}^{2}$

$= > f_{x} = \dfrac{ \delta(U)}{ \delta x}$

$= > f_{x} = \dfrac{ \delta(2 {x}^{3} - 4 {x}^{2} y + 6 {y}^{2} )}{ \delta x}$

$= > f_{x} = 6 {x}^{2} - 8xy + 0$

$= > f_{x} = 6 {x}^{2} - 8xy$

$= > f_{x} = 6 {(3)}^{2} - (8 \times 3 \times 2)$

$= > f_{x} = 54 - 48 = 6 \: \hat{i}$

$= > f_{x} = 54 - 48 = 6 \: \hat{i}$

For y axis:

$U = 2 {x}^{3} - 4 {x}^{2} y + 6 {y}^{2}$

$= > f_{y} = \dfrac{ \delta(U)}{ \delta y}$

$= > f_{y} = \dfrac{ \delta(2 {x}^{3} - 4 {x}^{2} y + 6 {y}^{2} )}{ \delta y}$

$= > f_{y} = 0 - 4 {x}^{2} + 12y$

$= > f_{y} = - 4 {x}^{2} + 12y$

$= > f_{y} = - 4 {(3)}^{2} +( 12 \times 2)$

$= > f_{y} = - 12 \: \hat{j}$

Hence total force:

$= > f_{net} = 6 \: \hat{i}- 12 \: \hat{j}$

Hence , A = 6 , B = -12.

So, final answer:

A + B = 6 + (-12) = -6