Question

A force parallel to the x-axis acts on a particle moving along the x-axis. This force produces a potential energy U(x) given by U=ax², where .a=1.20J/m⁴. What is the force (magnitude and direction) when the particle is at x = -0.800 m?

Answer 1

The force (magnitude and direction) acting on a particle at x = -0.800 m with a potential energy function U=ax², where a=1.20 J/m´, is 1.92 N in the positive x-direction.

Step-by-step explanation:

The force exerted on a particle at a position along the x-axis can be found by taking the negative gradient of the potential energy function, U(x). In this case, the potential energy is given by U=ax², with a constant a=1.20 J/m´. To find the force at a particular position x, we differentiate the potential energy function with respect to x and apply a negative sign.

The derivative of U=ax² with respect to x is dU/dx = 2ax. Therefore, the force F(x) acting on the particle is F(x) = -dU/dx = -2ax. When the particle is at x = -0.800 m, we plug this value into the formula to find the force, yielding F(-0.800 m) = -2(1.20 J/m´)(-0.800 m). Simplifying, we get F(-0.800 m) = 1.92 N. The positive sign indicates that the force acts in the positive x-direction.