Answer to Question #253948 in Physics for Kristine

Question

A force parallel to the 𝑥-axis acts on a particle moving along the 𝑥-axis. This force produces potential energy 𝑈(𝑥) given by 𝑈(𝑥) = 𝛼𝑥 4 , where 𝛼 = 0.630 J/m4 . What is the force (magnitude and direction) when the particle is at 𝑥 = −0.800 m?

Answer 1

Given:

$U(x)=\alpha x^4$

$\alpha=0.630 \:\rm J/m^4$

$x=-0.800\:\rm m$

The force

$$F=-\frac{dU}{dx}=-4\alpha x^3$$

$$F=-4*0.800*(-0.800)^3=1.64\:\rm N$$