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Physics 170 - Mechanics
Lecture 17
Potential Energy
1
Conservative Force: Gravity
Work done by gravity on a closed path is zero:
2
Nonconservative Force: Friction
Work done by friction on a closed path is not zero:
3
Conservative Forces and Paths
The work done by a conservative force is zero
on any closed path:
4
Conservative and
Nonconservative Forces
Conservative force: the work it does is stored in
the form of energy that can be released at a later
time
Examples of a conservative force: gravity. spring
Example of a nonconservative force: friction
Also: the work done by a conservative force in
moving an object around a closed path is zero;
this is not true for a nonconservative force.
5
Heat: Calories vs. joules
Heat energy is often produced by nonconservative forces.
Energy, particularly heat energy, is sometimes quantified in units of
calories, defined as the heat energy required to raise 1 g (= 1 ml = 1
cm3) of water by 1 degree Celsius. 1 cal = 4.186 J.
The equivalent SI unit is the kilocalorie, denoted by Cal or kcal,
which is the heat energy required to raise 1 kg (or 1 liter) of water
by 1 degree Celsius. 1 Cal = 1 kcal = 4.186 x 103 J. This is the unit
often used to indicate the energy content of food, and is measured
by burning the food in a calorimeter and measuring the heat output.
6
Conservative Forces
& Potential Energy
A potential energy can be associated with
any conservative force.
 Potential energy is an energy of position. The system has a unique value of
potential energy when the object is at A, a different value when the object is
at B, etc. Thus, the net change in potential energy is ΔU=UB-UA, and is the
same whether the object moves form A to B along Path 1 or along Path 2.
 Potential energy is transformed to kinetic energy with ΔK=-ΔU.
is independent of path, and ΔK is independent of path.
ΔU
 The change in the particle’s kinetic energy is related to the amount of work
done on the particle by force F. According to the work-kinetic-energy
theorem, ΔK=W. Because ΔK is independent of path, the work W done by the
force in moving from A to B must also be independent of path.
7
Work Done by
Conservative Forces: Gravity
Gravitational potential energy:
8
Work Done by
Conservative Forces: Gravity
If we pick up a ball and put it on the shelf, we
have done work on the ball. We can get that
energy back if the ball falls back off the shelf; in
the meantime, we say the energy is stored as
potential energy.
9
Example: A Falling Bottle
A 0.350 kg bottle falls from rest from a shelf that is 1.75 m
above the floor.
(a) Find the potential energy of the bottle when the bottle is on
the shelf.
(b) Find the kinetic energy of the bottle just before impact
with the floor.
=
(yf =0)
10
Example: Pike’s Peak or Bust
An 82.0 kg mountain climber is in the final
stage of the ascent of Pike’s Peak, which
4,301 m above sea level.
(a) What is the change in gravitational
potential energy as the climber gains the
last 100.0 m of altitude? Use U=0 at sea
level.
(b) Do the same calculation with U=0 at the
top of the peak.
11
Example: A Mountain Bar
A candy bar called the Mountain Bar
has an energy content when metabolized
of 212 Cal = 212 kcal. This is equivalent to
8.87 x 105 J.
If an 81.0 kg mountain climber eats a
Mountain Bar and magically converts all of
its energy content into gravitational
potential energy, how much altitude Δy
should he be able to gain?
12
Conservative Forces
and Potential Energy
A potential energy can be associated
with any conservative force.
Both are location-dependent and reversible potential energies.
Note that friction is not a conservative force and is irreversible.
13
Example:
The Speed of a Falling Rock
A rock of 1.0 Kg is released from rest.
Use both Betty’s and Bill’s perspectives to
calculate its speed just before it hits the
ground.
14
Example: The Speed of a Sled
Christine runs forward with her sled at
2.0 m/s. She hops onto the sled at the top
of a 5.0 m high, very slippery slope.
What is her speed at the bottom?
K1 + Ug1 = K0 + Ug0
½mv12+mgy1 = ½mv02+mgy0
v1 = [v02 + 2gy1]½
= [(2.0 m/s)2+2(9.80 m/s2)(5.0 m)]½
= 10.1 m/s
Notice that the steepness of the slope
and/or whether it has bumps and dips does not
matter in determining the answer. Only the
change in height and the initial speed are
relevant to the answer.
15