Question

A small block with a mass of 0.0400 kg is moving in the 𝑥𝑦-plane. The net force on the block is described by the potential energy function 𝑈(𝑥) = (5.80 𝐽 𝑚2 ⁄ )𝑥 2 − (3.60 𝐽 𝑚3 ⁄ )𝑦 3 . What are the magnitude and direction of the acceleration of the block when it is at the point (𝑥 = 0.300𝑚, 𝑦 = 0.600𝑚)? 

Answer 1

Find the energy at the given points:

$$U_x(0.3)=5.8·0.3^2=0.522\text{ J},\\ U_y(0.6)=3.6·0.6^3=0.778\text{ J}.$$

The force can be fund as

$$F=U/r.$$

The acceleration is

$$a=F/m=U(rm),\\\space\\ a_x=U_x/(xm)=0.522/(0.3·0.04)=43.5\text{ m/s}^2,\\ a_y=0.778/(0.6·0.04))=32.4\text{ m/s}^2.\\ a=\sqrt{a_x^2+a_y^2}=54.2\text{ m/s}^2,\\\space\\ \theta=\arctan\frac{a_y}{a_x}=36.7°$$

from positive x-direction to positive y.