Objectives
1. Identify several forms of energy.
2. Calculate kinetic energy for an object.
3. Apply the work-kinetic energy theorem to
solve problems.
4. Distinguish between kinetic and potential
energy.
5. Classify different types of potential energy.
6. Calculate the potential energy associated
with an object’s position.
Kinetic energy
• Kinetic energy is the energy of motion.
• An object which has motion - whether it be vertical
or horizontal motion - has kinetic energy.
• The equation for kinetic energy is:
KE = ½ mv2
– Where KE is kinetic energy, in joules
– v is the speed of the object, in m/s
– m is the mass of the object, in kg
• Kinetic energy is a scalar quantity.
Questions
1. Which of the following has kinetic energy?
a. a falling sky diver
b. a parked car
c. a shark chasing a fish
d. a calculator sitting on a desk
2. If a bowling ball and a volleyball are traveling at the
same speed, do they have the same kinetic energy?
3. Car A and car B are identical and are traveling at the
same speed. Car A is going north while car B is going
east. Which car has greater kinetic energy?
Kinetic Energy depends on mass and speed
• KE = ½ m v 2
• The equation shows that . . .
• the more mass a body has
• or the faster it’s moving
• . . . the more kinetic energy it has.
• Speed has more effect on Kinetic energy
than mass.
• KE is directly proportional to m, so doubling the mass doubles
kinetic energy, and tripling the mass makes it three times
greater.
Kinetic energy
Kinetic energy
• KE is proportional to v 2, so doubling the speed quadruples
kinetic energy, and tripling the speed makes it nine times
greater.
speed
mass
Sample Problem 5B
• A 7.00 kg bowling ball moves at 3.00 m/s. How
much kinetic energy does the bowling ball
heave? How fast must a 2.45 g table-tennis
ball move in order to have the same kinetic
energy as the bowling ball? Is this speed
reasonable for a table-tennis ball?
Example 1
•
An object moving at a constant speed of 25
meters per second possesses 450 joules of
kinetic energy. What is the object's mass?
Known:
• KE = 450 J
• v = 25 m/s
Unknown:
• m = ? kg
Solve:
KE = ½ mv2
450 J = ½ (m)(25 m/s)2
m = 1.4 kg
example 2
•
a.
b.
c.
d.
A cart of mass m traveling at a speed v has
kinetic energy KE. If the mass of the cart is
doubled and its speed is halved, the kinetic
energy of the cart will be
half as great
twice as great
one-fourth as great
four times as great
Example 3
•
Which graph best represents the relationship
between the kinetic energy, KE, and the velocity of
an object accelerating in a straight line?
a
b
c
d
Class work
• Page 174 - #1-5
1.
2.
3.
4.
170 m/s
38.8 m/s
The bullet with the greater mass; 2 to 1
2.4 J, 9.6 J; the bullet with the greater speed; 1
to 4
5. 1600 kg
Work-kinetic energy theorem
• The net work done on an object is equal to the
change in the kinetic energy.
Wnet  Fnet x
Wnet  m ax
Wnet  KE
v f  vi  2ax
2
2
v f  vi
2
ax 
Wnet  m(
Wnet
2
2
2
2
v f  vi
2
)
1
1
2
2
 m vf  m vi
2
2
Practice Problem #1
• A 1000-kg car traveling with a speed of 25 m/s skids to a stop.
The car experiences an 8000 N force of friction. Determine
the stopping distance of the car.
Wnet = ∆KE = KEf - KEi
(-8000N) • d = -312 500 0 J
d = 39.1 m
Practice Problem #2
• At the end of the Shock Wave roller coaster ride, the 6000-kg
train of cars (includes passengers) is slowed from a speed of
20 m/s to a speed of 5 m/s over a distance of 20 meters.
Determine the braking force required to slow the train of cars
by this amount.
Wnet = ∆KE = KEf - KEi
• The above problems have one thing in common: there is a force
which does work over a distance in order to remove mechanical
energy from an object.
• The force acts opposite the object's motion and thus does
negative work which results in a loss of the object's total
amount of mechanical energy. In each situation, the work is
related to the kinetic energy change.
TMEi + Wext = TMEf
KEi + Wext = 0 J
½ •m•vi2 + F•d•cos(180o) = 0 J
F•d = ½ •m•vi2
d ~ vi2
Stopping distance is dependent upon the square of the
velocity.
Stopping distance and initial velocity
Wnet = ∆KE
Wnet = 0 - ½ •m•vi2
Ff•d = - ½ •m•vi2
• Ff = μFnorm = μmg
• - μmg•d = - ½ mvi2
• d = vi2 / 2μg
d ~ v i2
practice
(m/s)
Stopping Distance (m)
0 m/s
0
5 m/s
4m
10 m/s
4 m x 22 = 16 m
15 m/s
4 m x 32 = 36 m
20 m/s
4 m x 42 = 64 m
25 m/s
4 m x 52 = 100 m
Example 5C
• On a frozen pond, a person kicks a 10.0 kg
sled, giving it an initial speed of 2.2 m/s. How
far does the sled move if the coefficient of
kinetic friction between and sled and the ice is
0.10?
Class work
• Page 176 practice 5C: #1-5
1.
2.
3.
4.
5.
7.8 m
21 m
5.1 m
300 N
a.-190 J; b. -280 J; c. 750 J; d. 280 J; e. 7.6 m/s
Potential energy
• An object can store energy as the result of its
position. Potential energy is the stored energy
of position possessed by an object.
• Two form:
– Gravitational
– Elastic
Gravitational potential energy
• Gravitational potential energy is the energy stored in
. (height)
an object as the result of its vertical position
• The energy is stored as the result of the gravitational
attraction of the Earth for the object.
PEgrav  m  g  h
Gravitational attraction between Earth and
the object:
m: mass, in kilograms
g: acceleration due to gravity = 9.81 m/s2
height, in meters
GPE and gravity
• When an object falls, gravity does positive
work. Object loses GPE.
• When an object is raised, gravity does
negative work. Object gains GPE.
Wgrav  m g(hi  h f )  m g(h f  hi )
h  h f  hi
Wgrav  m gh
Change in GPE only depends on
change in height, not path
As long as the object starts and ends at the same
height, the object has the same change in GPE because
gravity does the same amount of work regardless of
which path is taken.
example
• The diagram shows points A, B, and C at or near Earth’s
surface. As a mass is moved from A to B, 100. joules of work
are done against gravity. What is the amount of work done
against gravity as an identical mass is moved from A to C?
100 J
As long as the object starts and ends at the same height, the object
has the same change in GPE because gravity does the same amount of
work regardless of which path is taken.
Gravitational Potential Energy is
relative:
To determine the gravitational
potential energy of an object, a
zero height position must first
be assigned. Typically, the
ground is considered to be a
position of zero height.
But, it doesn’t have to be:
– It could be relative to the
height above the lab table.
– It could be relative to the
bottom of a mountain
– It could be the lowest
position on a roller coaster
Unit of energy
• The unit of energy is the same as work: Joules
• 1 joule = 1 (kg)∙(m/s2)∙(m) = 1 Newton ∙ meter
• 1 joule = 1 (kg)∙(m2/s2)
Work and energy has the same unit
example
•
•
•
•
•
How much potential energy is gained by an
object with a mass of 2.00 kg that is lifted
from the floor to the top of 0.92 m high
table?
Known:
m = 2.00 kg
h = 0.92 m
g = 9.81 m/s2
• unknown:
• PE = ? J
Solve:
∆PE = mg∆h
∆PE = (2.00 kg)(9.81m/s2)(0.92 m)
= 18 J
•
The graph of
gravitational
potential energy
vs. vertical height
for an object near
Earth's surface
gives the weight of
the object.
The weight of the
object is the slope of
the line.
Weight = 25 J/1.0 m = 25 N
m = weight / g = 2.5 kg
• Springs are a special instance of a device that can
store elastic potential energy due to either
compression or stretching.
• A force is required to compress or stretch a spring;
the more compression/stretch there is, the more
force that is required to compress it further.
• For certain springs, the amount of force is directly
proportional to the amount of stretch or
compression (x); the constant of proportionality is
known as the spring constant (k).
Hooke’s Law
F = kx
Spring force = spring constant x displacement
• F in the force needed to displace (by stretching or
compressing) a spring x meters from the equilibrium
(relaxed) position. The SI unit of F is Newton.
• k is spring constant. It is a measure of stiffness of the
spring. The greater value of k means a stiffer spring
because more force is needed to stretch or compress
it that spring. The SI units of k are N/m.
• x the distance difference between the length of
stretched/compressed spring and its relaxed
(equilibrium) spring.
example
• Determine the x in F = kx
F = kx
force
• Spring force is directly proportional to the
elongation of the spring (displacement)
The slope represents spring
constant: k = F / x
elongation
caution
elongation
• Sometimes, we might see a graph such as this:
The slope represents the inverse of spring constant:
Slope = 1/k = x / F
force
example
• Given the following data table and corresponding graph,
calculate the spring constant of this spring.
example
• A 20.-newton weight is attached to a spring, causing
it to stretch, as shown in the diagram. What is the
spring constant of this spring?
example
•
The graph below shows elongation as a function of
the applied force for two springs, A and B.
Compared to the spring constant for spring A, the
spring constant for spring B is
1. smaller
2. larger
3. the same
Example 12A
• If a mass of 0.55 kg attached to a vertical
spring stretches the spring 2.0 cm from its
original equilibrium position, what is the
spring constant?
Class work
• Page 441, Practice 12A #1-4
1.
2.
3.
4.
a. 15 N/m; b. less stiff
320 N/m
2700 N/m
81 N
Lab 14 – Hooke’s Law (1)
1.
2.
3.
4.
5.
Purpose: To determine the spring constant of a given spring.
Material: spring, masses, meter stick.
Procedure: Hook different masses on the spring, record the
force Fs (mg) and corresponding elongation x. Plot the graph of
Force vs. elongation
Data section: should contain colomns: force applied,
elongation.
– Data measured directly from the experiment. The units of
measurements in a data table should be specified in
column heading only.
Data analysis: Graph force vs. elongation on graph paper,
answer following questions:
– What does the slope mean in Force vs. elongation graph?
– Determine the spring constant
Force (N)
Elongation (m)
Force (N)
Force vs. elongation
Elongation (m)
Elastic potential energy
• Elastic potential energy is the energy stored in
elastic materials as the result of their
stretching or compressing.
• Elastic potential energy can be stored in
– Rubber bands
– Bungee cores
– Springs
– trampolines
Elastic potential energy in a spring
•
Elastic potential energy is the Work done on the spring.
PEs  Favg  x
0kx
1
PE s  (
)  x  ( k  x)  x
2
2
1
PE s  k  x 2
2
•
•
k: spring constant
x: amount of compression or extension relative to
equilibrium position
Elastic potential energy
Elastic potential energy is directly
proportional to x2
elongation
Example 1
• As shown in the diagram, a 0.50-meter-long spring is
stretched from its equilibrium position to a length of 1.00
meter by a weight. If 15 joules of energy are stored in the
stretched spring, what is the value of the spring constant?
PE = ½ kx2
15 J = ½ k (0.50 m)2
k = 120 N/m
Example 2
• The unstretched spring in the diagram has a length of 0.40
meter and a spring constant k. A weight is hung from the
spring, causing it to stretch to a length of 0.60 meter. In terms
of k, how many joules of elastic potential energy are stored in
this stretched spring?
PEs = ½ kx2
PEs = ½ k(0.20 m)2
PEs = (0.020 k) J
Example 3
• Determine the potential energy stored in the spring with a
spring constant of 25.0 N/m when a force of 2.50 N is applied
to it.
Solve:
Given:
PEs = ½ k∙x2
Fs = 2.50 N
To find x, use Fs = kx,
k = 25.0 N/m
(2.50 N) = (25.0 N/m)(x)
Unknown:
x = 0.100 m
PEs = ? J
PEs = ½ (25.0 N/m)(0.100 m)2
PEs = 0.125 J
Example 4
• A 10.-newton force is required to hold a
stretched spring 0.20 meter from its rest
position. What is the potential energy stored
in the stretched spring?
Given:
F = 10. N
x = 0.20 m
Unknown: PEs = ? J
PEs = Favg∙x
PEs = (½ F)∙x
PEs = ½ (10. N)(.20 m)
PEs = 1.0 J
Sample Problem 5D
• A 70.0 kg stuntman is attached to a bungee cord with an
unstretched length of 15.0 m. He jumps off a bridge spanning
a river from a height of 50.0 m. When he finally stops, the
cord has a stretched length of 44.0 m. Treat the stuntman as a
point mass, and disregard the weight of the bungee cord.
Assuming the spring constant of the bungee cord is 71.8 N/m,
what is the total potential energy relative to the water when
the man stops falling?
Class work
• Page 180 – practice 5D #1-3
1. 3.3 J
2. 0.031 J
3. a. 785 J;
b. 105 J;
c. 0.00 J