The kinetic energy of the mass at the instant it passes back through its equilibrium position is about 1.20 J

<h3>Further explanation</h3>
Let's recall Elastic Potential Energy formula as follows:

where:
<em>Ep = elastic potential energy ( J )</em>
<em>k = spring constant ( N/m )</em>
<em>x = spring extension ( compression ) ( m )</em>
Let us now tackle the problem!

<u>Given:</u>
mass of object = m = 1.25 kg
initial extension = x = 0.0275 m
final extension = x' = 0.0735 - 0.0275 = 0.0460 m
<u>Asked:</u>
kinetic energy = Ek = ?
<u>Solution:</u>
<em>Firstly , we will calculate the spring constant by using </em><em>Hooke's Law</em><em> as follows:</em>






<em>Next , we will use </em><em>Conservation of Energy</em><em> formula to solve this problem:</em>







<h3>Learn more</h3>

<h3>Answer details</h3>
Grade: High School
Subject: Physics
Chapter: Elasticity
Looks like you need to review through the lesson and take notes as it tells you in the lesson what each of these are.
Answer:
S = V t + 1/2 a t^2 = 5 m/s * 5 s + .2 m/s^2 * 25 s^2 =
25 m + 5 m = 30 m distance traveled
Vf = V + a t = 5 m/s + .4 m/s^2 * 5 s = (5 + 2) m/s = 7 m/s final velocity
Yes that's correct. Also zeros in between non-zero numbers are significant figures