Conservation of Energy
And Potential Energy
Law of Conservation of Energy
The system is isolated and
allows no exchange with the
environment.
No mass can enter or leave
No energy can enter or leave
Energy is constant, or
conserved
E = U + K + Eint
= Constant
Law of Conservation of Mechanical Energy
We only allow U
and K to
interchange
We ignore Eint
(thermal energy)
E=U+K
= Constant
Law of Conservation of
Mechanical Energy
E=U+K=C
or
for gravity
  Ug = mghf - mghi
  K = ½ mvf2 - ½ mvi2
for springs
  Us = ½ kxf2 - ½ kxi2
  K = ½ mvf2 - ½ mvi2
E = U + K = 0
Pendulum Energy
h
½mv12 + mgh1 = ½mv22 + mgh2
For any points two points in the pendulum’s swing
Spring Energy
0
m
½ kx12 + ½ mv12
= ½ kx22 + ½ mv22
-x
m
For any two points in a spring’s
oscillation
m
x
Conservative Forces
Conservative Forces
The done on a particle
moving b/w any two
points is independent of
the path taken
The work done on a
particle moving through
a closed path is zero
Ex: gravity
Wc  
xf
xi
Fc dx   U
W=Ui – Uf = -ΔU
This is only true of
conservative forces.
This happens because
the force is negative (Fg
= -mgy or Fx= -kx) and
thus the change in
energy (work)ends up
being –Uf – Ui or –ΔU.
Practice Problem:
A 2.0 m pendulum is released from rest when the support string is
at an angle of 25 degrees with the vertical. What is the speed
of the bob at the bottom of the string?
q
Lcosq
L
h = L – Lcosq
h = 2-2cosq
h = 0.187 m
h
1.35 m/s = v
Practice Problem:
A single conservative force of F = (3i + 5j) N acts on a 4.0
kg particle. Calculate the work done if the particle if the
moves from the origin to r = (2i - 3j) m.
a. Does the result depend on path?
b. What is the speed of the particle at r if the speed at the
origin was 4.0 m/s?
c. What is the change in potential energy of the system?
W = -9J
a) no, Wc is independent of path
b) 3.4 m/s
c) 9J
Practice Problem:
A bead slides on the loop-the-loop shown and is released
from height h = 3.5 R.What is the speed at point A?
(assume all energy is conserved)
VA=√(3gR)
Non-conservative forces
Nonconservative Forces
The work done on a
particle b/w any two points
is dependent of the path
taken
Causes a change in
mechanical energy (the
sum of the kinetic and
potential energies)
Ex: friction and drag
 Wtot = Wnc + Wc = ΔK
 Wnc = ΔK – Wc
(Wc = -ΔU)
 Wnc = ΔK + ΔU
 (Δkfriction= -Fkd – this is
the energy lost due to
friction, the internal
energy that goes into
the object (as thermal
energy))
Practice Problem:
A 2,000 kg car starts from rest and coasts down from the top of
a 5.00 m long driveway that is sloped at an angel of 20o with
the horizontal. If an average friction force of 4,000 N impedes
the motion of the car, find the speed of the car at the bottom
of the driveway. (remember this is a nonconservative force)
Vf = 3.7 m/s
Practice Problem:
A parachutist of mass 50 kg jumps out of a hot air balloon 1,000 meters
above the ground and lands on the ground with a speed of 5.00
m/s. How much energy was lost to friction during the descent?
4.9 x 105 J
Force and Potential Energy
 Before we discuss the relationships between potential energy
and force, lets review a couple of relationships.
 Wc = Fx (if force is constant)
 Wc =  Fdx = - dU = -U (if force varies)
  Fdx = - dU
 Fdx = -dU
 F = -dU/dx
Energy Diagrams: Stable Equilibrium
U
Stable Equilibrium:
Any displacement from
equilibrium results in a
force directed back
towards x = 0
The positions of stable
equilibrium correspond
to the points where U(x)
is a minimum
Example:
A spring
A ball in a bowl
-x
x
x
1st derivative:
minimum gives position of stable equilibrium
2nd derivative:
would give the spring constant
x and –x give the turning points, a spring will
oscillate b/w these points because it can’t
exceed ½kx2
Energy Diagrams: Stable Equilibrium
A spring in stable equilibrium:
Fs = -dUs/dx = -kx
The force is thus equal to the
negative of the slope of the
energy curve.
Us= ½kx2
Us
If a force stretches the spring, x is
+ and the slope is +, thus F is - and
brings the spring back to
equilibrium.
If a force compresses the spring, x
is – and the slope is -, thus F is +
and brings the spring back to
equilibrium.
x
Energy Diagrams: Unstable Equilibrium
Unstable Equilibrium:
Any displacement from
equilibrium results in an
acceleration away from
that point
U
The positions of unstable
equilibrium correspond
to the points where U(x)
is a maximum
Example:
A pencil balanced
vertically
x
Energy Diagrams: Neutral Equilibrium
Neutral Equilibrium:
Any displacement from
equilibrium results in
neither a restorative nor
a disruptive force
U
Example: A ball on a flat
table
x
Practice Problem:
The potential energy associated with the force between two
neutral atoms in a molecule is modeled by the Lennard-Jones potential
energy function:
Where x is the distance b/w the atoms, and σ and ε are determined
experimentally. In this case σ= .263 nm and ε= 1.51 x 10-22 J.
We expect to find the stable equilibrium point where the potential
energy of the system is at a minimum, find the equilibrium separation of
the two atoms.
X = 2.95 x 10-10 m
Molecular potential energy diagrams
Graph of the potential energy curve for the molecule at
various distances between the atoms
The potential energy is
quite large when the atoms
are very close together, at
a minimum when they are
at their critical separation,
and increases again when
the atoms move apart.
When U is a minimum, the
atoms are in stable
equilibrium. This is the most
likely separation between
the atoms
U
x