Question

You stretch a spring by a distance of 0.05 m. The spring has a spring constant of 125 N/m. When you release the spring, it snaps back. What is the kinetic energy of the spring as it reaches its natural length?

Answer 1

In this problem energy is conserved means Energy before = Energy after, so the Elastic Potential Energy is being transformed into Kinetic Energy and when it reach its natural length, the energy that is left is only Kinetic Energy. So we can assume E.P.E = K.E approximately 0.15625 J

Answer 2

KE = 0.16 J

Step-by-step explanation:

As we know that total energy is always remains conserved

so here we can say that initial potential energy stored in the spring is equal to the kinetic energy of the object when it comes to relaxed state

So here we have

`U = (1)/(2)kx^2`

here we know that

`k = 125 N/m`

x = 0.05 m

now from above equation

`U = (1)/(2)(125)(0.05)^2`

`U = 0.16 J`

so total potential energy here is same as the final kinetic energy which is 0.16 J