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Conservation of Energy:
Matilda Berke
5/2/17
Background
The Law of Conservation of Energy states that energy cannot be created or destroyed. This
means that the total energy in an isolated system doesn’t change over time. We can use this
principle to determine how energy ​transforms ​from form to form throughout the various systems
we observe. It’s especially useful in situations where we’re trying to determine the kinetic,
potential, and internal energies of objects at specific points in time.
Major Topics
Energy transfer (potential energy, kinetic energy, internal energy)
Work
Conservative forces
Non-conservative forces
Power
Important Vocabulary
WORK, the exertion of a force over a distance, transfers energy into a system
CONSERVATIVE FORCES can be “stored” as potential energy and are available for use as
kinetic energy later (e.g. gravity, electricity)
NON-CONSERVATIVE FORCES are not forms of “stored” energy because they’re not
available for use as kinetic energy (e.g. friction)
MECHANICAL WAVES transfer energy through disturbances in surrounding media (e.g. ocean
waves, seismic waves)
HEAT is energy transfer driven by temperature difference, often related to internal energy
INTERNAL ENERGY is usually used to describe the result of frictional forces on an object
MATTER TRANSFER is when the displacement of matter across boundaries physically
transfers energy
ELECTRICAL TRANSMISSION involves the transfer of energy via electric currents
ELECTROMAGNETIC RADIATION refers to the transfer of energy through electromagnetic
waves (e.g. light, microwaves)
POWER is the rate of energy transfer within a given quantity
Important Equations
b
W = ∫ F dx or W = F dcosθ : work is equal to force integrated over distance
P =
a
W
t
: power is equal to work done over time
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ΔEsystem = ΣT : total energy of the system is equal to the sum of the the energy transfers within
the system, expands to…
ΔK + ΔU + ΔEint = W + Q + T mw + T mt + T et + T er : the sum of kinetic energy, potential
energy, and internal energy is equal to the sum of work, heat transfer, mechanical waves, matter
transfer, electrical transmission, and electromagnetic radiation
W net = ΔK : net work is equal to the change in kinetic energy
K = 12 mv 2 : kinetic energy of an object is equal to one-half the mass times velocity squared
U = mgh : gravitational potential energy is equal to mass times gravitational acceleration times
distance from “ground”
K = 12 kx2 : elastic potential energy is equal to one-half the spring constant times displacement
squared
Examples
1. A roller coaster car of mass 1000 kg follows this fun two-hill path. Assuming it starts from
rest at the top of the tallest hill (the point indicated with the dashed outline), how fast is it going
at the top of the second hill? Discount friction.
Because gravity is the only force acting here, we’re in a conservative system. This means that we
can really just calculate the work done in moving the roller coaster car the vertical difference
between the tops of the two hills here:
b
W = ∫ F dx → W = F x
a
F = ma
W = mgx → W = mg(1000 − 700)
W net = ΔK → 300mg = Δ 12 mv 2
The car is at rest at the top of the first hill, so v​initial​ = 0 making K​initial​ = 0 as well:
300mg = 12 mv 2
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A mass is provided in the problem, but we don’t need it! Yay! Cancel the masses and solve for v:
300(9.8) = 12 v 2 → ​v = 76.68 m/s
2. Someone pushes a 0.5 kg block along a frictionless table with a speed of 3 m/s until it’s
stopped by a spring (k = 100 N/m). Determine how much the spring compresses.
Let’s start with:
b
W = ∫ F dx
a
When we integrate this quantity, we get:
W =− 12 kx2
So work for a spring:
W = ΔK → W = Δ 12 mv 2
Since the block starts from rest, its v​initial =
​ 0 so its K​initial =
​ 0
− 12 mv 2 =− 12 kx2 → x = v
√
m
k
√
→ x=3
0.5
100
→ ​x = 0.21 m
3. In a thrilling turn of events, Jamal (playing Dr. Frederick Frankenstein) slips up and pushes
Mateo (70 kg, playing young Frankenstein’s monster) across a 5 meter long lab table with a
moderate slope of 20 degrees. If Mateo’s initial velocity is 0.5 m/s and his final velocity is 0.3
m/s, how much work is done to slow his slide?
HINT: Use friction!
Right when Mateo gets pushed:
ΣE = K + U
U = mgh → U = (70)(9.8)(5sin20) → U = 1173.13 J
K = 12 mv 2 → K = 12 (70)(0.5)2 → K = 8.75 J
ΣE initial​ = 1173.13 + 8.75 = 1181.88 J
At the bottom of the table:
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U=0
K = 12 mv 2 → K = 12 (70)(0.3)2 → K = 3.15 J
We know we have a difference in energies from the top to the bottom, so let’s use that to find out
how much work friction did to slow Mateo down:
1181.88 + W = 3.15 → W = ​-1178.73 J
From here, we can potentially use what we already know to do a lot with this: we can use a force
analysis/free-body diagram to determine the coefficient of friction, we can figure out what the
frictional force is, etc…
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