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
<h2>
Speed with which it return to its initial level is 100 m/s</h2>
Explanation:
We have equation of motion v² = u² + 2as
Initial velocity, u = 100 m/s
Acceleration, a = -9.81 m/s²
Final velocity, v = ?
Displacement, s = 0 m
Substituting
v² = u² + 2as
v² = 100² + 2 x -9.81 x 0
v² = 100²
v = ±100 m/s
+100 m/s is initial velocity and -100 m/s is final velocity.
Speed with which it return to its initial level is 100 m/s
Answer:
the answer for the question is the last option