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Lecture 16
General Physics (PHY 2130)
• Energy
Conservative and non-conservative forces Elastic potential energy Power
http://www.physics.wayne.edu/~apetrov/PHY2130/
Lightning Review
Last lecture: 1. Work and energy:
work: connection between forces and energy
potential energy Review Problem: Tarzan swings on a 30.0-m-long vine
initially inclined at an angle of 37.0° with the vertical. What is his speed at the bottom of the swing (a) if he starts from rest? (b) if he pushes off with a speed of 4.00 m/s?
Review problem
Tarzan swings on a 30.0-m-long vine initially inclined at an angle of 37.0° with the vertical. What is his speed at the bottom of the swing (a) if he starts from rest? (b) if he pushes off with a speed of 4.00 m/s?
yi
Work Done by Varying Forces
• Recall: the work done by a variable force acting on an object that undergoes a displacement is equal to the area under the graph of F versus x
5
x (m)
Fx (N) F3
F2
F1
x3 x2 x1
The work done by F1 is ( )0111 −= xFW
Example: What is the work done by the variable force shown below?
The net work is then W1+W2+W3.
The work done by F2 is ( )1222 xxFW −=The work done by F3 is ( )2333 xxFW −=
Potential Energy Stored in a Spring • Involves the spring constant (or force constant), k
• Hooke’s Law gives the force
• F = - k x
• F is the restoring force • F is in the opposite direction of x • k depends on how the spring was formed, the
material it is made from, thickness of the wire, etc.
7
Example: (a) If forces of 5.0 N applied to each end of a spring cause the spring to stretch 3.5 cm from its relaxed length, how far does a force of 7.0 N cause the same spring to stretch? (b) What is the spring constant of this spring?
(a) For springs F∝x. This allows us to write .2
1
2
1
xx
FF
=
Solving for x2: ( ) cm. 9.4cm 5.3N 5.0N 0.7
11
22 =⎟
⎠
⎞⎜⎝
⎛== xFFx
N/cm. 43.1cm 3.5N 0.5
1
1 ===xFk
(b) What is the spring constant of this spring? Use Hooke’s law:
N/cm. 43.1cm 4.9N 0.7
2
2 ===xFkOr
8
Example: An ideal spring has k = 20.0 N/m. What is the amount of work done (by an external agent) to stretch the spring 0.40 m from its relaxed length?
Fx (N)
x (m) x1=0.4 m
kx1
( )( ) ( )( ) J 6.1m 4.0N/m 0.2021
21
21
curveunder Area22
111 ====
=
kxxkx
W
Potential Energy in a Spring • Elastic Potential Energy
• related to the work required to compress a spring from its equilibrium position to some final, arbitrary, position x
2
21 kxPEs =
( )
.21
20:
220
2,1cos
:,cos
2
0
xkxkxWThus
kxFFFF
butxFW
spr
xx
spr
=−
−=
−=
+=
+==
=
θ
θ
This is called elastic potential energy:
Conservation of Energy including a Spring • If needed, the PE of the spring is added to both sides of
the conservation of energy equation
•
fsgisg PEPEKEPEPEKE )()( ++=++
11
Example: A box of mass 0.25 kg slides along a horizontal, frictionless surface with a speed of 3.0 m/s. The box encounters a spring with k = 200 N/m. How far is the spring compressed when the box is brought to rest?
m 11.0
021
210 22
=⎟⎟⎠
⎞⎜⎜⎝
⎛=
+=+
+=+
=
vkmx
kxmv
KUKUEE
ffii
fi
Given: m = 0.25 kg k = 200 N/m vi = 3.0 m/s vf = 0 m/s Find:
x = ?
Idea: we are given velocity, mass and spring constant. Let’s use conservation of energy: kinetic energy pf the box was transformed into elastic potential energy of the spring.
Nonconservative Forces with Energy Considerations
• When nonconservative forces are present, the total mechanical energy of the system is not constant
• The work done by all nonconservative forces acting on parts of a system equals the change in the mechanical energy of the system
EnergyW ativenonconserv Δ=
Nonconservative Forces and Energy • In equation form:
• The energy can either cross a boundary or the energy is transformed into a form not yet accounted for
• Friction is an example of a nonconservative force
)()(
)(
iiffnc
fiifnc
PEKEPEKEWorPEPEKEKEW
+−+=
−−−=
Transferring Energy
• Mechanical Waves • a disturbance propagates
through a medium • Examples include sound,
water, seismic
Transferring Energy
• Electromagnetic radiation • any form of electromagnetic
waves • Light, microwaves, radio waves
Problem Solving with Nonconservative Forces
• Define the system • Write expressions for the total initial and final energies • Set the Wnc equal to the difference between the final and
initial total energy • Follow the general rules for solving Conservation of
Energy problems
Power • Often also interested in the rate at which the energy transfer takes place
• Power is defined as this rate of energy transfer •
• SI units are Watts (W)
•
vFtWP ==
2
2
smkg
sJW •==
Power, cont. • US Customary units are generally hp (horsepower)
• need a conversion factor
• Can define units of work or energy in terms of units of power: • kilowatt hours (kWh) are often used in electric bills
W746slbft550hp1 ==
22
Example: A race car with a mass of 500.0 kg completes a quarter-mile (402 m) race in a time of 4.2 s starting from rest. The car’s final speed is 125 m/s. What is the engine’s average power output? Neglect friction and air resistance.
watts103.921
5
2
av
×=Δ
=Δ
Δ=
Δ
Δ+Δ=
Δ
Δ=
t
mv
tK
tKU
tEP
f
Given: m=500.0 kg s = 402 m t = 4.2 s vf = 125 m/s vi = 0 m/s Find:
P = ?
Idea: to compute power we need energy and time. Time is given, while energy (kinetic) can be computed
Notice that the distance information was not needed.
Center of Mass
• The point in the body at which all the mass may be considered to be concentrated • When using mechanical energy, the change in potential energy is
related to the change in height of the center of mass
Momentum • From Newton’s laws: force must be present to change an
object’s velocity (speed and/or direction) Wish to consider effects of collisions and corresponding
change in velocity
Method to describe is to use concept of linear momentum
×
scalar vector
Linear momentum = product of mass velocity
Golf ball initially at rest, so some of the KE of club transferred to provide motion of golf ball and its change in velocity
Momentum
• Vector quantity, the direction of the momentum is the same as the velocity’s
• Applies to two-dimensional motion as well
yyxx mvpandmvp ==
vmp =
Size of momentum: depends upon mass depends upon velocity
Impulse
• In order to change the momentum of an object (say, golf ball), a force must be applied
• The time rate of change of momentum of an object is equal to the net force acting on it
•
• Gives an alternative statement of Newton’s second law • (F Δt) is defined as the impulse • Impulse is a vector quantity, the direction is the same as
the direction of the force
tFporamtvvm
tpF net
ifnet Δ=Δ=
Δ
−=
Δ
Δ= :)(
Graphical Interpretation of Impulse
• Usually force is not constant, but time-dependent
• If the force is not constant,
use the average force applied • The average force can be
thought of as the constant force that would give the same impulse to the object in the time interval as the actual time-varying force gives in the interval
( )i
i it
impulse F t area under F t curveΔ
= Δ =∑
If force is constant: impulse = F Δt
Example: Impulse Applied to Auto Collisions • The most important factor is the collision time or the time it takes the person to come to a rest • This will reduce the chance of dying in a car crash
• Ways to increase the time • Seat belts • Air bags
The air bag increases the time of the collision and absorbs some of the energy from the body
ConcepTest
Suppose a ping-pong ball and a bowling ball are rolling toward you. Both have the same momentum, and you exert the same force to stop each. How do the time intervals to stop them compare?
1. It takes less time to stop the ping-pong ball. 2. Both take the same time. 3. It takes more time to stop the ping-pong ball.
ConcepTest
Suppose a ping-pong ball and a bowling ball are rolling toward you. Both have the same momentum, and you exert the same force to stop each. How do the time intervals to stop them compare?
1. It takes less time to stop the ping-pong ball. 2. Both take the same time. 3. It takes more time to stop the ping-pong ball.
Note: Because force equals the time rate of change of momentum, the two balls loose momentum at the same rate. If both balls initially had the same momenta, it takes the same amount of time to stop them.
Problem: Teeing Off
A 50-g golf ball at rest is hit by “Big Bertha” club with 500-g mass. After the collision, golf leaves with velocity of 50 m/s.
a) Find impulse imparted to ball b) Assuming club in contact with
ball for 0.5 ms, find average force acting on golf ball
Problem: teeing off
Given: mass: m=50 g = 0.050 kg velocity: v=50 m/s Find: impulse=? Faverage=?
1. Use impulse-momentum relation:
2. Having found impulse, find the average force from the definition of impulse:
( )( )smkg
smkgmvmvpimpulse if
⋅=
−=
−=Δ=
50.2050050.0
Nssmkg
tpFthustFp
3
3
1000.5105.0
50.2,
×=
×
⋅=
Δ
Δ=Δ⋅=Δ
−
Note: according to Newton’s 3rd law, that is also a reaction force to club hitting the ball:
( )iiff
ifif
R
VMvmVMvm
orVMVMvmvm
ortFtF
+=+
−−=−
Δ⋅−=Δ⋅
,
, of club
CONSERVATION OF MOMENTUM
Conservation of Momentum • Definition: an isolated system is the one that has no external forces acting on it
• A collision may be the result of physical contact between two objects
• “Contact” may also arise from the electrostatic interactions of the electrons in the surface atoms of the bodies
Momentum in an isolated system in which a collision occurs is conserved (regardless of the nature of the forces between the objects)
Conservation of Momentum
The principle of conservation of momentum states when no external forces act on a system consisting of two objects that collide with each other, the total momentum of the system before the collision is equal to the total momentum of the system after the collision
Conservation of Momentum • Mathematically:
• Momentum is conserved for the system of objects • The system includes all the objects interacting with each other • Assumes only internal forces are acting during the collision • Can be generalized to any number of objects
ffii vmvmvmvm 22112211 +=+
Problem: Teeing Off (cont.)
Let’s go back to our golf ball and club problem:
( )( ) sm
kgsmkgvv
smkgvvm
smvmsmkgp
if
if
55.0
50.2so,50.2:Club
50gramm50,50.2:Ball
−=⋅−
=−
⋅−=−
=Δ
=⋅=Δ
factor of 10 times smaller
ConcepTest
Suppose a person jumps on the surface of Earth. The Earth
1. will not move at all 2. will recoil in the opposite direction with tiny velocity 3. might recoil, but there is not enough information provided to see if that could happened
ConcepTest
Suppose a person jumps on the surface of Earth. The Earth
1. will not move at all 2. will recoil in the opposite direction with tiny velocity 3. might recoil, but there is not enough information provided to see if that could happened
Note: momentum is conserved. Let’s estimate Earth’s velocity after a jump by a 80-kg person. Suppose that initial speed of the jump is 4 m/s, then:
smkgsmkgV
smkgVMpsmkgp
Earth
EarthEarth
2324 103.5
106320
so,320:Earth320:Person
−×−=×
⋅−=
⋅−==Δ
⋅=Δ
tiny negligible velocity, in opposite direction