An ideal spring hangs from the ceiling. a 1.25-kg mass is hung from the spring. after all vibrations have died away, the spring is found to have stretched 0.0275 m from its original length. the mass is then pushed up 0.0735 m from its equilibrium position and released. what is the kinetic energy of the mass at the instant it passes back through its equilibrium position?

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

Let's recall Elastic Potential Energy formula as follows:

where:

Ep = elastic potential energy ( J )

k = spring constant ( N/m )

x = spring extension ( compression ) ( m )

Let us now tackle the problem!

Given:

mass of object = m = 1.25 kg

initial extension = x = 0.0275 m

final extension = x' = 0.0735 - 0.0275 = 0.0460 m

Asked:

kinetic energy = Ek = ?

Solution:

Firstly , we will calculate the spring constant by using Hooke's Law as follows:

Next , we will use Conservation of Energy formula to solve this problem:

Grade: High School

Subject: Physics

Chapter: Elasticity

KE= 1.20 J

Explanation:

Given that

m = 1.25 kg

x= 0.0275  m

A= 0.0735 m

F= K x = m g

k = mg/x

k=445.9 N/m

This is the spring constant.

Kinetic energy of mass when it passes through equilibrium given as

By putting the values

KE= 1.20 J