Single – Theory
Objective:
Date: ___________
Work Done by Gravitational Potential Energy
Homework: Assignment (1-23)
Do PROBS #’s (59) Ch. 6 +
Do 1997 #1, AP 1999 #1, (handout)
AP Physics “B”
Mr. Mirro
Date: ________
Gravitational Potential Energy and Transfer of Energy
The GRAVITATIONAL force is a well-known force that can do positive (+) or negative (-) work.
For example, a ball of mass (m) moving vertically downward
only has the force (Fg) of GRAVITY acting on it !
The initial height (h0) of the ball and the final height of the ball (hF)
are both measured with respect to (wrt) the earths surface so that the
change in height (Δh), represents the displacement (s) of the object.
Since the displacement (s) and the weight (mg) act downward, the
work done by the ball on the ground (+W) can be expressed as:
W = Fg * sy cos θ
1
W = mg h cos 0° ⇒
+mgh
Similarly, the work done in lifting the ball against gravity (-W) through the same distance (s) can be
expressed as:
-1
W = mg h cos 180° ⇒ - m g h
To obtain an expression for the gravitational potential energy, lets analyze the change in work done as the
object falls.
Wgravity = Wf - Wi = mg (hf) - mg (h0) ⇒
mg Δh
This equation indicates that the work done by the gravitational force is equal to the difference between the
initial and final values of the quantity mgh.
Gravitational Potential Energy
The gravitational potential energy PE is the energy that an object of mass m has
by virtue of its position relative to the surface of the earth. That position is
measured by the height h of the object relative to an arbitrary zero level:
PE = mgh
Energy is neither created nor destroyed – it simply transfers into different forms of energy. A simple way
to visualize the transfer of energy as an object falls is with an “energy bucket.”
At each point in the fall of the object, the amount of kinetic energy
and potential energy are shown as if they were two differently
colored materials in the “bucket.”
v0 = 0
y = 3.0 m
1
kg
PE
As the object falls from rest it looses “height” and thus, looses some of
its potential energy due to gravity. However, as it falls over time –
it gains “speed”, thereby gaining energy of motion or kinetic energy.
Since the total amount of material in the energy-bucket, total mechanical
energy, remains constant, the gravitational potential energy lost (PELost)
is equal to the kinetic energy gained (KEGained).
That is,
KE
y = 1.5 m
PE
PELost = KEGained
y=0m
vf = 7.7 m/s
Thus, the speed of the object at the midway point can be inferred by considering
the portion of the potential energy that was transferred into kinetic energy.
PELost = mg Δh = (1 kg)(10m/s2)[1.5m] = 15 J
So,
KEGained = 15 J
as well !
Therefore, the speed can be computed by using the fact that KE = ½ mv2.
15 J = ½ (1 kg) v2
v =
√
2(15)
1
=
5.47 m/s
KE
AP Physics “B”
Mr. Mirro
Date: _________
Gravitational Potential Energy and Transfer of Energy
Ex 1: A 2 kg pendulum bob is attached to a frictionless pivot through a massless taught arm that is two
meters long. At the top of its arc the pendulum has speed v = 0 as it changes direction through each
cycle. [CutnellP6.32sim]
a. What is the speed of the pendulum at the instant it crosses its
equilibrium position ?
b. What is the period of this pendulum ?
Ex 2: A gymnast springs vertically upward from a trampoline as shown. The gymnast leaves the
trampoline at a height of 1.2 m and reaches a maximum height of 4.8 m before falling back down.
All heights are measured with respect to the ground. Ignoring air resistance, determine the initial
speed v0 with which the gymnast leaves the trampoline. [Cutnell6.7]
Ex 3: A 1000 kg roller-coaster moves from point A to point B and then to point C as shown.
[Giancoli6.7]
10 m
B
A
0
a. What is its gravitational potential energy at point B
relative to point A ?
C
15m
b. What is its gravitational potential energy at point C relative to point A ?
c. What is the change in potential energy when it goes from point B to point C ?
d. What kinetic energy has been gained by the time the coaster crosses point C ?
e. How fast could the coaster be traveling at the instant it approaches point C ?