Download Lecture 16 - Wayne State University Physics and Astronomy

General Physics (PHY 2130)
Lecture 16
•  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
Example: What is the work done by the variable force shown below?
Fx (N)
F3
F2
F1
x1
x2
x3
x (m)
The work done by F1 is
W1 = F1 (x1 − 0)
The work done by F2 is
W2 = F2 (x2 − x1 )
The work done by F3 is
W3 = F3 (x3 − x2 )
The net work is then W1+W2+W3.
Potential Energy Stored in a Spring
•  Involves the spring constant (or force
constant), k
•  Hooke’s Law gives the force
• 
F=-kx
•  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
Solving for x2:
F1 x1
= .
F2 x2
F2
⎛ 7.0 N ⎞
x2 =
x1 = ⎜
⎟(3.5 cm ) = 4.9 cm.
F1
⎝ 5.0 N ⎠
(b) What is the spring constant of this spring? Use Hooke’s law:
F1 5.0 N
k= =
= 1.43 N/cm.
x1 3.5 cm
Or
F2 7.0 N
k=
=
= 1.43 N/cm.
x2 4.9 cm
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)
kx1
x1=0.4 m
x (m)
W = Area under curve
1
1 2 1
2
= (kx1 )(x1 ) = kx1 = (20.0 N/m)(0.4 m ) = 1.6 J
2
2
2
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
Wspr = (F cosθ ) x, but :
F0 + Fx 0 + Fx − kx
=
=
2
2
2
− kx
1
= 0−
x = k x2.
2
2
cosθ = 1, F =
Thus : Wspr
This is called elastic potential energy:
1 2
PEs = kx
2
Conservation of Energy including a Spring
•  If needed, the PE of the spring is added to both sides of
the conservation of energy equation
• 
( KE + PE g + PEs )i = ( KE + PE g + PEs ) f
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?
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.
Ei = E f
U i + Ki = U f + K f
1 2 1 2
0 + mv = kx + 0
2
2
⎛ m ⎞
⎟v = 0.11 m
x = ⎜⎜
⎟
k
⎝
⎠
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
Wnonconservative = ΔEnergy
Nonconservative Forces and Energy
•  In equation form:
Wnc = KE f − KEi − ( PEi − PE f ) or
Wnc = ( KE f + PE f ) − ( KEi + PEi )
•  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
Transferring Energy
•  By Work
•  By applying a force
•  Produces a displacement
of the system
Transferring Energy
•  Heat
•  The process of
transferring heat by
collisions between
molecules
Transferring Energy
•  Mechanical Waves
•  a disturbance propagates
through a medium
•  Examples include sound,
water, seismic
Transferring Energy
•  Electrical transmission
•  transfer by means of
electrical current
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
• 
W
P=
= Fv
t
•  SI units are Watts (W)
• 
J kg • m2
W= =
s
s2
Power, cont.
•  US Customary units are generally hp (horsepower)
•  need a conversion factor
ft lb
1 hp = 550
= 746 W
s
•  Can define units of work or energy in terms of units of power:
•  kilowatt hours (kWh) are often used in electric bills
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.
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
ΔE ΔU + ΔK
Pav =
=
Δt
Δt
1 2
mv
ΔK 2 f
=
=
= 9.3 ×10 5 watts
Δt
Δt
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 and Collisions
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
Golf ball initially at rest, so
some of the KE of club
transferred to provide motion
of golf ball and its change in
velocity
 Method to describe is to use concept of linear momentum
Linear momentum = product of mass
scalar
× velocity
vector
Momentum
p = mv
•  Vector quantity, the direction of the momentum is the
same as the velocity’s
•  Applies to two-dimensional motion as well
p x = mv x and p y = mv y
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
• 
F net
Δ p m(v f − v i )
=
=
= ma or : Δ p = F net Δt
Δt
Δt
•  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
Graphical Interpretation of Impulse
•  Usually force is not constant,
but time-dependent
impulse = ∑ Fi Δti = area under F (t ) curve
Δti
•  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
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
1. Use impulse-momentum relation:
Given:
impulse = Δp = mv f − mvi
= (0.050 kg )(50 m s ) − 0
mass: m=50 g
= 0.050 kg
velocity: v=50 m/s
Find:
impulse=?
Faverage=?
= 2.50 kg ⋅ m s
 
2. Having found impulse, find the average
force from the definition of impulse:
Δp 2.50 kg ⋅ m s
=
Δt
0.5 ×10 −3 s
= 5.00 ×10 3 N
Δp = F ⋅ Δt , thus F =
 
Note: according to Newton’s 3rd law, that is also a reaction force to club hitting the ball:
of club
F ⋅ Δt = − F R ⋅ Δt , or
(
)
mv f − mv i = − M V f − M V i , or
mv f + M V f = mv i + M V i
CONSERVATION OF MOMENTUM
Conservation of Momentum
•  Definition: an isolated system is the one that has
no external forces acting on it
Momentum in an isolated system in which a
collision occurs is conserved (regardless of the
nature of the forces between the objects)
•  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
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:
m1 v1i + m2 v 2i = m1 v1 f + m2 v 2 f
•  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
Problem: Teeing Off (cont.)
Let’s go back to our golf ball and club problem:
Ball : Δp = 2.50 kg ⋅ m s , m = 50 gramm
Δv = 50 m s
(
(
)
)
Club : m v f − v i = −2.50 kg ⋅ m s , so
− 2.50 kg ⋅ m s
v f − vi =
= −5m s
0.5 kg
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:
Person : Δp = 320 kg ⋅ m s
Earth : Δp = M EarthVEarth = −320 kg ⋅ m s , so
VEarth =
− 320 kg ⋅ m s
− 23
=
−
5
.
3
×
10
m s
6 ×10 24 kg
tiny negligible velocity, in opposite direction