Find the magnitude of the force at the position x = 2.0 m and y = 3.0 m associated with the potential energy function U = (7.0 J)+ (5.0 J/m)x + (4.0 J/m)y.

Question

Quizplus · Accepted Answer

The force $\vec{F}$ associated with a potential energy function $U(x, y)$ is given by $\vec{F} = -\nabla U$, where $\nabla U$ is the gradient of $U$. For $U = 7.0 + 5.0x + 4.0y$, the components of the force are $F_x = -\frac{\partial U}{\partial x} = -5.0 \, \text{N}$ and $F_y = -\frac{\partial U}{\partial y} = -4.0 \, \text{N}$. The magnitude of the force is $\sqrt{F_x^2 + F_y^2} = \sqrt{(-5.0)^2 + (-4.0)^2} = \sqrt{25 + 16} = \sqrt{41} \approx 6.4 \, \text{N}$.

Find the Magnitude of the Force at the Position X