# Lesson 09 - MnE - Kinetic and Potential Energy

```Mechanical Energy Forms:
Kinetic and Potential
Energy
Lesson 9
Announcements
HW #8 due today
 HW #9 due tomorrow


Lunch Bunch – Waves #2 this week

Waves HW #1 due Tuesday
HW Quiz TUESDAY
 Optics Exam repair
Friday 7am, 4th, 5th, 3pm

AP Physics B Standards
Mechanical Energy forms: Kinetic and Potential Energy
LESSON 9:

I.C.1. Work and the Work-Energy Theorem
b)Students should understand and be able to apply the work-energy
theorem, so they can:
(1) Calculate the change in kinetic energy or speed that results from performing a
specified amount of work on an object.
(2) Calculate the work performed by the net force, or by each of the forces that make
up the net force, on an object that undergoes a specified change in speed or kinetic
energy.
(3) Apply the theorem to determine the change in an object’s kinetic energy and speed
that results from the application of specified forces, or to determine the force that is
required in order to bring an object to rest in a specified distance.
Lesson Objectives
Students will be able to
1.
2.
distinguish between conservative and non-conservative
forces
apply the work-energy theorem to calculate changes in
kinetic energy due to net external forces.
Force types
Forces acting on a system can be
divided into two types according to
how they affect potential energy.
 Conservative forces can be related
to potential energy changes.
 Non-conservative forces cannot be
related to potential energy changes.
 So, how exactly do we distinguish
between these two types of forces?

Conservative forces

Work is path independent.



Work along a closed path is zero.



If the starting and ending points are the same, no work is
done by the force.
Work changes potential energy.
Examples:



Work can be calculated from the starting and ending
points only.
The actual path is ignored in calculations.
Weight
Spring force
Conservation of mechanical energy holds!
More on paths and
conservative forces.
Q: Assume a conservative force
moves an object along the
various paths. Which two works
are equal?
A:
W2 = W3
(path independence)
Q: Which two works, when added
together, give a sum of zero?
A:
W1 + W2 = 0
or
W1 + W3 = 0
(work along a closed path is zero)
Non-conservative forces

Work is path dependent.




Work along a closed path is NOT zero.
Work changes mechanical energy.
Examples:



Knowing the starting and ending points is
not sufficient to calculate the work.
Friction
Drag (air resistance)
Conservation of mechanical energy
does not hold!
The Work-Energy Theorem
 The
net work due to all forces equals
the change in the kinetic energy of a
system.
 Wnet = K
 Wnet:
work due to all forces acting on an
object
 K: change in kinetic energy (Kf – Ki)
Sample Problem: A 15-g acorn falls from a tree and lands on the ground
10.0 m below with a speed of 11.0 m/s.
a) What would the speed of the acorn have been if there had been no air
resistance?
b)
Did air resistance do positive, negative or zero work on the acorn? Why?
Sample Problem: A 15-g acorn falls from a tree and lands on the ground
10.0 m below with a speed of 11.0 m/s.
c) How much work was done by air resistance?
d)
What was the average force of air resistance?
Conservative forces and
Potential energy
Potential energy is energy of position or
configuration. Sometimes called “stored”
energy
 For gravity: Ug = mgh

m: mass
 g: acceleration due to gravity
 h: height above the “zero” point


For springs: Us = ½ k x2
k: spring force constant
 x: displacement from equilibrium position

Conservative forces and
Potential energy

Conservative forces change the potential energy of
a system.
Wc  U


If a conservative force does positive work on a
system, potential energy is lost.
If a conservative force does negative work, potential
energy is gained.
Sample Problem:
A vertical spring stores 0.962 J in spring potential energy when a 3.0 kg mass
is suspended from it.
a) By what factor does the spring potential energy change if the mass
attached to the spring is doubled?
Us = ½ k x2, which seems mass independent. But mass changes the displacement!
Fs – mg = 0
k x = mg
x = mg/k
If m is doubled, x is doubled,
and since Us is proportional to x2, the energy is quadrupled.
b) Verify your answer to part (a) by calculating the spring potential energy
when a 6.0 kg mass is attached to the spring.
1) Solve for k
Us = ½ k x2
Us = ½ k (mg/k)2
0.962 J = ½ k (3.0 kg*9.8/k)2
2*0.962 /(3*9.8)2 = k/k2
k = 450 N/m
2) Solve for Us
Us = ½ k x2
Us = ½ k (mg/k)2
Us = ½ (450 N/m)[(6.0 kg * 9.8 m/s2 / 450 N/m)2
Us = 3.842 J
3) Compare to initial Us
3.842/4 = 0.960, which is close enough to 0.962
Sample Problem:
A candy bar called the Mountain Bar has a calorie content of 210.0 Cal =
210 kCal, which is equivalent to 8.791 x 105 J. If an 82.0 kg mountain climber
eats a Mountain Bar and magically converts it all to potential energy, what
gain of altitude would be possible?
Sample Problem:
As an Acapulco cliff diver drops to the water from a height of 40.0 m, his
gravitational potential energy decreases by 25,000 J. How much does the
diver weigh?
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