Two 8.00-kilogram masses are
connected to each other by a spring with a force constant of 25.0 newtons per meter
and a rest length of 1.00 meters. The spring has been compressed to
0.900 meters in length. The masses travel toward each
other. And each mass moves at 0.500 meters
per second. What is the total energy in the
We can label the total system
energy we want to solve for capital 𝐸 and begin by drawing a sketch of the
situation. In this scenario, we have two
identical masses — we’ve labeled them 𝑚 — of value 8.00 kilograms, connected by a
spring of spring constant 𝑘. The spring started out at a length
we’ve called 𝐿 sub 𝑖 for its initial length 1.00 meters. And now it’s at a length we’ve
called 𝐿 sub 𝑓, 0.900 meters. The two masses move towards one
another at speeds 𝑣, where 𝑣 is given as 0.500 meters per second. Knowing all this, we want to solve
for the total energy in the system at this moment in time.
We know that this total energy will
be equal to the sum of the system’s kinetic plus its potential energy. We recall that kinetic energy, in
general, is equal to one-half an object’s mass times its speed squared. And since we have two identical
masses, 𝑚, moving at identical speeds, 𝑣, we can rewrite this expression for
kinetic energy simply as 𝑚 times 𝑣 squared.
When it comes to the potential
energy of our system, we have a spring which has been compressed from its natural
length. The potential energy due to an
elastic deformation is equal to one-half a spring’s constant, 𝑘, multiplied by its
displacement from equilibrium, Δ𝑥, squared. In our case, we can write this
elastic potential energy as one-half the force constant, 𝑘, times the quantity 𝐿
sub 𝑖 minus 𝐿 sub 𝑓 squared.
In the problem statement, we’re
given 𝑚, 𝑣, 𝑘, 𝐿 sub 𝑖, and 𝐿 sub 𝑓. So we’re ready to plug in and solve
for 𝐸. When we enter in the values for 𝑚,
𝑣, 𝑘, 𝐿 sub 𝑖, and 𝐿 sub 𝑓 and enter this expression on our calculator, we
find that, to three significant figures, 𝐸 is 2.13 joules. That’s the total energy of the
system, kinetic plus potential.