Question

simple harmonic motion
when you substitute in the physical variables for the area, height, and base represented in the force-displacement graph, the equation for the area of a triangle, ( a = \frac{1}{2}bh ), becomes ( pe_s = \frac{1}{2}fd ). substituting in hooke’s law and using ( d = x ) allows us to determine the energy in the spring based on the spring’s constant and the spring’s displacement ( pe_s = \frac{1}{2}kx^2 ).
switch back to the “systems” section of the simulation. set “applied force” to 50.0 n, “top spring” to 600 n/m, and “bottom spring” to 400 n/m.
determine the amount of energy stored in the system. enter your answer in the box.

Explanation:

Step1: Find displacement using Hooke's law

Hooke's law is \( F = k_{eq}x \), where \( k_{eq} \) for springs in parallel is \( k_1 + k_2 \). Here, \( k_1 = 600\ N/m \), \( k_2 = 400\ N/m \), so \( k_{eq}=600 + 400 = 1000\ N/m \). Given \( F = 50.0\ N \), solve for \( x \):
\( x=\frac{F}{k_{eq}}=\frac{50.0}{1000}=0.05\ m \)

Step2: Calculate total potential energy

The formula for elastic potential energy in a spring system is \( PE_s=\frac{1}{2}k_{eq}x^2 \). Substitute \( k_{eq}=1000\ N/m \) and \( x = 0.05\ m \):
\( PE_s=\frac{1}{2}\times1000\times(0.05)^2 \)
\( PE_s = 500\times0.0025 = 1.25\ J \)

Answer:

\( 1.25\) (in joules)