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Consul'ng room for Phys 201 office hours TA Office hours are in 2131 Chamberlin
link to schedule from course info page
Chapter 6 Work and Energy
October 8 and October 13, 2009
Work and Energy Energy is one of the most important concepts in physics Alterna've approach to analyze some mechanics situa'ons Ini'ally considered the energy of mo'on (kine'c energy) But many other forms (with names)
Many applica'ons beyond mechanics Thermodynamics (movement of heat) Quantum mechanics... Chemistry Biology ……………….(every science)
Very useful tool in understanding our physical world. You will learn new (some'mes much easier) ways to solve problems.
Energy and Newton’s Laws
The concept of energy turns out to be even more general than Newton’s Laws.
But the importance of mechanical energy in classical mechanics is a consequence of Newton’s laws.
Defini'on of Work: Constant Force
θFr
Vector “Dot Product”
�
W = F ⋅ Δ r = FΔr cosθ = FrΔr
Work is a scalar.
Defini'on of Work….
Only the component of F parallel to the displacement is doing work.
θ
F cos θ
Ques'on
You li[ an object 1 m from the floor to a shelf. The work done by the earth’s gravita'onal field on the object is: A. posi've B. zero C. nega've
Ques'on
You li[ an object 1 m from the floor to a shelf. The work done by the earth’s gravita'onal field on the object is: A. posi've B. zero C. nega've
The object’s displacement is opposite in direction to the force of gravity on the object. So the work done by the gravitational force on the object is negative.
correct
Ques'on
You push a heavy box 1m across a rough floor. Which statements are correct?
A. You have done posi've work on the box. B. You have done nega've work on the box. C. The earth’s gravita'onal force has done nega've work on the box.
E. You have done nega've work on the outside world.
Ques'on
You push a heavy box 1m across a rough floor. Which statements are correct?
A. You have done posi've work on the box. B. You have done nega've work on the box. C. The earth’s gravita'onal force has done nega've work on the box.
E. You have done nega've work on the outside world.
correct
Defini'on of Work
Total work, W, of a force F ac'ng
through a displacement Δx = dx î is:
Work = Area under F(x) curve
Fx
x Δxi
Fi
�
W = F i ⋅ Δ x i
i∑
�
= Fxdxx1
x2∫
Work done by a spring force
Force exerted by a spring extended a distance x is:
Fx = ‐kx (Hooke’s law)
Force done by spring on an object when the displacement of spring is changed from xi to xf is:
Defini'on of Kine'c Energy
Kine'c energy (K.E.) of a par'cle of mass m moving with speed v is defined as
K.E. = ½mv2
Kine'c energy is a useful concept because of the Work/Kine'c Energy theorem, which relates the work done on an object to the change in kine'c energy.
First do some simple examples, then show general relationship between work and kinetic energy.
Work/Kine'c Energy Theorem: Special case of constant force
v1 v2
m
Work/Kine'c Energy Theorem: Special case of constant force
v1 v2
m
Work/Kine'c Energy Theorem: Special case of constant force
v1 v2
m
Problem: Work and Energy
Two blocks have masses m1 and m2, where m1 > m2. They are sliding on a fric'onless floor and have the same kine'c energy when they encounter a long rough stretch with μ>0, which slows them down to a stop.
Which one will go farther before stopping?
(A) m1 (B) m2 (C) They will go the same distance
m1
m2
Problem: Work and Energy
Two blocks have masses m1 and m2, where m1 > m2. They are sliding on a fric'onless floor and have the same kine'c energy when they encounter a long rough stretch with μ>0, which slows them down to a stop.
Which one will go farther before stopping?
(A) m1 (B) m2 (C) They will go the same distance
m1
m2
Solution method on following slides
Problem: Work and Energy (Solu'on)
The work‐kine'c energy theorem says that for any object,
WNET = ΔK. In this example, the only force that does work is fric'on (since both N and mg are perpendicular to the block’s mo'on).
m f
N
mg
Problem: Work and Energy (Solu'on)
The net work done to stop a box is ‐fD = ‐μmgD.
The work‐kine'c energy theorem says that for any object, WNET = ΔK, so WNET=Kf‐Ki=0‐Ki.
Since the boxes start out with the same kine'c energy, we have
μm1gD1=μm2gD2 and D1/D2=m2/m1. Since m1>m2, we must have D2>D1.
m1
D1
m2
D2
A falling object
What is the speed of an object that starts at rest and then falls a ver'cal distance H?
H
v0 = 0
v
A falling object
What is the speed of an object that starts at rest and then falls a ver'cal distance H?
Work done by gravita'onal force WG = FΔr = mgH
Work/Kine'c Energy Theorem:
�
WG = mgH = 12 mv2
⇒ v = 2gH
H
v0 = 0
v
The scalar product (or dot product)
θab
θ
ba
Examples of dot products
Suppose Then
x
y
z
Proper'es of dot products
ax
ay
More proper'es of dot products
Which of the statements below is correct?
A. The scalar product of two vectors can be nega've.
B. AcB=c(BA), where c is a constant.
C. The scalar product can be non‐zero even if two of the three components of the two vectors are equal to zero.
E. All of the above statements are correct.
Which of the statements below is correct?
A. The scalar product of two vectors can be nega've.
B. AcB=c(BA), where c is a constant.
C. The scalar product can be non‐zero even if two of the three components of the two vectors are equal to zero.
E. All of the above statements are correct.
Work/KE Theorem (Variable Force, 3‐d)
�
W = F ⋅d r r 1
r 2∫
�
= m d v
dt⋅d r
r (t i )
r (tf )∫
Work done by a force exerted over a path is
�
= m d v
dt⋅ d r
dtdt
ti
tf∫ = m d
v dt
⋅ v dtti
tf∫
�
= m ddt12⋅v2⎛
⎝ ⎜
⎞ ⎠ ⎟ dt
ti
tf∫
�
= 12mvf
2 − 12mvi
2
�
ΔW = 12mvf
2 − 12mvi
2
Net work done on object = change in kinetic energy of object
What about mul'ple forces?
Comments about work
W = F Δr
• The 'me interval over which the force acts is not relevant
• No work is done if: F = 0 or Δr = 0 or F is perpendicular to Δr
More comments about work
W = F Δr
No work is done if F and Δr are perpendicular.
No work is done by tension T
No work is done by normal force N.
Ques'on
The work needed to maintain uniform circular mo'on in a horizontal plane is: A. zero B. posi've and constant C. nega've and constant
E. nega've and variable
Ques'on
The work needed to maintain uniform circular mo'on in a horizontal plane is: A. zero B. posi've and constant C. nega've and constant
E. nega've and variable
Force applied (towards the center of the circle) is perpendicular to the direction of motion at all points in the uniform circular motion.
Power
Defini'on of power:
Power is the rate at which a force does work.
Power
�
P = dWdt
�
dW = F ⋅d r =
F ⋅ v dt
⇒ P = dWdt
= F ⋅ v
Unit of power: 1 watt = 1 joule/second
Ques'on
A sports car accelerates from zero to 30 mph in 1.5 s. How long does it take for it to accelerate from zero to 60 mph, assuming the power of the engine to be independent of velocity and neglec'ng fric'on?
Ques'on
A sports car accelerates from zero to 30 mph in 1.5 s. How long does it take for it to accelerate from zero to 60 mph, assuming the power of the engine to be independent of velocity and neglec'ng fric'on?
Power is constant, so , a constant.
Integra'ng with respect to 'me, and no'ng that the ini'al velocity is zero, one gets . So genng to twice the speed takes 4 'mes as long, and the 'me to reach 60 mph is 4×1.5 = 6s.
�
ddt
12 mv
2( ) = C
�
12 mv
2 = Ct
Ques'on A cart on an air track is moving at 0.5 m/s when the air is suddenly turned off. The cart comes to rest a[er traveling 1 m. The experiment is repeated, but now the cart is moving at 1 m/s when the air is turned off. How far does the cart travel before coming to rest?
Ques'on A cart on an air track is moving at 0.5 m/s when the air is suddenly turned off. The cart comes to rest a[er traveling 1 m. The experiment is repeated, but now the cart is moving at 1 m/s when the air is turned off. How far does the cart travel before coming to rest?
Work done by fric'on force over a distance d is ‐Fd. If ini'al velocity is v, then work/kine'c energy theorem says that the stopping distance D is determined by
FD=½mv2. If the ini'al velocity is doubled, then the stopping distance goes up by a factor of 4. So the stopping distance for the larger ini'al velocity is 4 m/s.
Power P is required to li[ a body a distance d at constant speed v. What power is required to li[ the body a distance 2d at constant speed 3v?
A. P B. 2P C. 3P
E. 3P/2
Power P is required to li[ a body a distance d at constant speed v. What power is required to li[ the body a distance 2d at constant speed 3v?
A. P B. 2P C. 3P
E. 3P/2
[Power = Fv; force needed does not change (since speed is still constant).]
If a fighter jet doubles its speed, by what factor should the power from the engine change? (Assume that the drag force on the plane is propor'onal to the square of the plane’s speed.)
A. by half B. unchanged C. doubled
E. 8 'mes
If a fighter jet doubles its speed, by what factor should the power from the engine change? (Assume that the drag force on the plane is propor'onal to the square of the plane’s speed.)
A. by half B. unchanged C. doubled
E. 8 ?mes
Magnitude of power is Fv. When the velocity v is doubled, the drag force goes up by a factor of 4, and Fv goes up by a factor of 8
Center‐Of‐Mass Work For systems of particles that are not all moving at the same velocity, there is a work-kinetic energy relation for the center of mass.
So
where is the translational kinetic energy
Net work done on collection of objects
change in translational kinetic energy of system =
Problem (6‐62)
The magnitude of the single force ac'ng on a par'cle of mass m is given by F = bx2, where b is a constant. The par'cle starts from rest at x=0. A[er it travels a distance L, determine its (a) kine'c energy and (b) speed.
Problem (6‐62)
The magnitude of the single force ac'ng on a par'cle of mass m is given by F = bx2, where b is a constant. The par'cle starts from rest at x=0. A[er it travels a distance L, determine its (a) kine'c energy and (b) speed.
�
W = F ⋅d x
0
L∫ = bx2dx
0
L∫ = b
3L3work done by force:
So (a) kinetic energy is K=bL3/3
(b) Since ½mv2=bL3/3,
�
v = 2bL3
3m
Ques'on
A 1kg block slides 4 m down a fric'onless plane inclined at 30 degrees to the horizontal. What is the speed of the block as it leaves the inclined plane?
30°
Ques'on
A 1kg block slides 4 m down a fric'onless plane inclined at 30 degrees to the horizontal. What is the speed of the block as it leaves the inclined plane?
30°
No friction, so only force that does work is the gravitational force. Work done by gravity is W = mgL sinθ. By Work/Kinetic Energy Theorem, we have
mgL sinθ = ½mv2, so v = (2gL sinθ)1/2 = 6.3 m/s
A person pushes against a spring, the opposite end of which is arached to a fixed wall. The spring compresses. Is the work done by the spring on the person posi've, nega've or zero?
A. posi've B. nega've C. zero
A person pushes against a spring, the opposite end of which is arached to a fixed wall. The spring compresses. Is the work done by the spring on the person posi've, nega've or zero?
A. posi've B. nega?ve C. zero
A light rope runs through two fric'onless pulleys of negligible mass. A mass, m, is hung from one of the
pulleys and a force, F, is applied to one end of the rope so that the mass moves at a constant speed. What is the force needed to move the mass at a constant speed?
A. 0 B. ½ mg C. mg
E. 4mg
A light rope runs through two fric'onless pulleys of negligible mass. A mass, m, is hung from one of the
pulleys and a force, F, is applied to one end of the rope so that the mass moves at a constant speed. What is the force needed to move the mass at a constant speed?
A. 0 B. ½ mg C. mg
E. 4mg