Answer to Question #253952 in Physics for Kristine

Question #253952

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)? 


1
Expert's answer
2021-10-31T18:11:08-0400

Find the energy at the given points:


"U_x(0.3)=5.8\u00b70.3^2=0.522\\text{ J},\\\\\nU_y(0.6)=3.6\u00b70.6^3=0.778\\text{ J}."

The force can be fund as


"F=U\/r."

The acceleration is


"a=F\/m=U(rm),\\\\\\space\\\\\na_x=U_x\/(xm)=0.522\/(0.3\u00b70.04)=43.5\\text{ m\/s}^2,\\\\\na_y=0.778\/(0.6\u00b70.04))=32.4\\text{ m\/s}^2.\\\\\na=\\sqrt{a_x^2+a_y^2}=54.2\\text{ m\/s}^2,\\\\\\space\\\\\n\\theta=\\arctan\\frac{a_y}{a_x}=36.7\u00b0"

from positive x-direction to positive y.


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