### Video Transcript

A string has a force constant of
53.0 newtons per meter. An object, initially at rest, with
a mass of 0.960 kilograms is suspended from it. The object descends, stretching the
spring, oscillates, and then comes to rest. How much is the spring stretched
when the object has come to rest after oscillating? Calculate the decrease in the
gravitational potential energy of the object between its position at the point at
which it is attached to the unextended spring and its position at the point at which
it comes to rest after oscillating. Calculate the energy stored in the
spring by its extension.

In this three-part exercise, we
first wanna solve for the displacement of the spring after the mass has been hung on
it and comes to rest. Next, we want to solve for the
change in gravitational potential energy of the mass between the two points from
which it is hung on the unextended spring and the point at which it comes to
rest. And finally, we wanna calculate the
energy stored in the spring by its extension. Weโll call this PE sub ๐ .

In this scenario, we start off with
a vertically oriented spring at its natural, or equilibrium, length. The spring, weโre told, has spring
constant ๐ of 53.0 newtons per meter. Then, we take a mass, weโve called
it ๐, of value 0.960 kilograms and attach it to the free end of the spring. In response to this, the spring
stretches, oscillates, and then comes to rest at a displacement weโve called ฮ๐ฅ
from its original position.

To solve for ฮ๐ฅ, letโs start by
considering the forces that are acting on the mass ๐. When the mass has come to rest
after oscillating on the end of the spring, we know that the gravitational force
acting on the mass is equal in magnitude to the spring force on the mass. So, we can write ๐ times ๐ is
equal to the spring constant ๐ times ฮ๐ฅ. Rearranging this expression, we see
that ฮ๐ฅ is equal to ๐ times ๐ over ๐, where ๐ weโll treat as exactly 9.8 meters
per second squared.

Plugging in for these three values,
when we calculate ฮ๐ฅ, we find that, to three significant figures, itโs 17.8
centimeters. Thatโs the amount the spring is
stretched because of the mass ๐. Now, letโs move on to solving for
the change in gravitational potential energy of the mass ๐ over this
displacement.

Recalling that, in general,
gravitational potential energy is equal to an objectโs mass times the acceleration
due to gravity times its height relative to some reference level. Since weโre calculating the change
in gravitational potential energy of the mass, we can write that itโs equal to ๐
times ๐ times ฮ๐ฅ. No negative sign needed.

When we plug in for these three
values, being careful to write our displacement ฮ๐ฅ in units of meters, we find that
ฮPE sub ๐ is 1.67 joules. Thatโs the change in the massโ
gravitational potential energy. Finally, we want to solve for the
potential energy stored up in the spring as result of this extension.

We recall that in general that
potential energy is equal to half a springโs constant multiplied by its displacement
from equilibrium squared. So, we can write that PE sub ๐ is
equal to one-half ๐ times ฮ๐ฅ squared, where ๐ the spring constant is known and
ฮ๐ฅ we solved for earlier. Substituting in for these values
using a distance in meters for our displacement ฮ๐ฅ, we find that PE sub ๐ is 0.835
joules. Thatโs how much potential energy
the spring has gained by virtue of being extended.

We may wonder why the spring
potential energy is not equal to the change in gravitational potential energy. Recall that, before coming to rest,
our mass ๐ oscillated up and down on the spring. During that up-and-down motion, it
lost energy through friction with the air as well as through the spring. That helps account for the
difference between these two values.