Download Chapter 6 Work and Energy Definimon of Work: Constant Force

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Defini5on of Work: Constant Force Chapter 6 Work and Energy Work (dot product)
Work by gravity, by spring
Kinetic energy, power
Work-kinetic energy theorem
C.M. system
Vector
“Dot Product”
Work is a scalar.
Units are Newton meter: Joule = N x m
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More on “dot product” (or scalar product) Work is done in li<ing the box (you do +work; gravity ‐) No work is done on the bucket when held s5ll, or to move horizontally ba
θ
θ
ab
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Examples of dot products More proper5es of dot products y
x
z
Suppose
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Then
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What about mul5ple forces? Which of the statements below is correct? A.  The scalar product of two vectors can be nega5ve. B.  AcB=c(BA), 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. 2/20/11
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FN
You are towing a car up a hill with constant velocity. The work done on the car by the normal force is: 1. posi5ve 2. nega5ve 3. zero correct
T
Normal force is perpendicular to displacement
cos θ = 0
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FN
V
T
T is pointing in the direction of motion therefore, work done by this force is positive.
W
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You are towing a car up a hill with constant velocity. The work done on the car by the gravita5onal force is: 1. posi5ve correct
2. nega5ve 3. zero Phys 201, Spring 2011
W
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FN
You are towing a car up a hill with constant velocity. The total work done on the car by all forces is: 1. posi5ve 2. nega5ve correct
3. zero Phys 201, Spring 2011
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T
Constant velocity implies that there is no net force
acting on the car, so there is no work being done overall
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T
There is a non-zero component of gravitational
force pointing opposite the direction of motion.
W
You are towing a car up a hill with constant velocity. The work done on the car by the tension force is: correct
1. posi5ve 2. nega5ve 3. zero 2/20/11
V
W
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Li<ing an object: Work by a variable force: A 3000–kg truck is to be loaded onto a ship by a crane that exerts an upward
force of 31 kN on the truck. This force, which is strong enough to overcome the
gravitational force and keep the truck moving upward, is applied over a distance
of 2.0 m. Find
(a) the work done on the truck by the crane,
(b) the work done on the truck by gravity,
(c) the net work done on the truck.
Total work, W, of a force F ac5ng through a displacement Δx = dx î is an integra5on: Fx
(a) the work done by the crane:
Wc = Fc d = 31 kN x 2 m = 62 kJ
(b) the work done by gravity:
Wg = Fg d = (-31 kN) x 2 m = -62 kJ
Fi
x
(c) the net work done on the truck:
W = Wc + Wg = 0
Δxi
Work = Area under F(x) curve
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Work done by a spring force Force exerted by a spring extended a distance x is: Fx = ‐kx (Hooke’s law) 2/20/11
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A person pushes against a spring, the opposite end of which is a^ached to a fixed wall. The spring compresses. Is the work done by the spring on the person posi5ve, nega5ve or zero? Force done by spring on an object when the displacement of spring is changed from xi to xf is: 2/20/11
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A.  posi5ve B.  nega9ve C.  zero 2/20/11
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Energy is that quality of a substance or object which “causes something to happen”; or “capability of exer5ng forces”; or “ability to do work”… The “energy of mo5on”. Work done on the object increases its energy,
-- by how much? (i.e. how to calculate the value?)
W=Fd
The vagueness of the defini5on is due to the fact that energy can result in many effects. Phys 201, Spring 2011
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Defini5on of Kine5c Energy Kine5c energy (K.E.) of a par5cle of mass m moving with speed v is defined as a=
2
W=m
It is convertible into other forms without loss
(i.e it is conserved)
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W = (ma) d
 Electrical Energy
 Chemical Energy
 Mechanical Energy
 Nuclear Energy
K.E. = ½mv2 Kine5c energy is a useful concept because of the Work/Kine5c Energy theorem, which relates the work done on an object to the change in kine5c energy. 1
2
2
V = V 02+ 2 a d
F = ma
2d
2
V - V0
2d
2
V - V0
d
1
2
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A falling object What is the speed of an object that starts at rest and then falls a ver5cal distance H? Work done by gravita5onal force WG = FΔr = mgH v0 = 0
H
Work/Kine5c Energy Theorem: v
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A skier of mass 50 kg is moving at speed 10 m/s at point P1 down a ski slope with negligible fric5on. What is the skier’s kine5c energy when she is at point P2, 20 m below P1? A. 2500 J B. 9800 J C. 12300 J D. 13100 J E. 15000 J Energy and Newton’s Laws The importance of mechanical energy in classical mechanics is a consequence of Newton’s laws. The concept of energy turns out to be even more
general than Newton’s Laws.
Work-kinetic energy theorem:
KEf – KEi = Wg = mgH
KEf = ½×50×102 + 50×9.8×20 = 2500 J + 9800 J
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Problem: Work and Energy Problem: Work and Energy (Solu5on) Two blocks have masses m1 and m2, where m1 > m2. They are sliding on a fric5onless floor and have the same kine5c 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 The net work done to stop a box is ‐fD = ‐μmgD. The work‐kine5c energy theorem says that for any object, WNET = ΔK so WNET=Kf‐Ki=0‐Ki. Since the boxes start out with the same kine5c energy, we have μm1gD1=μm2gD2 and D1/D2=m2/m1. Since m1>m2, we must have D2>D1. Solution next page 
m1
m1
m2
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D1
m2
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Problem: Power: P: Work done by unit 5me. The magnitude of the single force ac5ng on a par5cle of mass m is given by F = bx2, where b is a constant. The par5cle starts from rest at x=0. A<er it travels a distance L, determine its (a) kine5c energy and (b) speed. work done by force:
Units: Watt
So (a) kinetic energy is K=bL3/3
(b) Since ½mv2=bL3/3,
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Ques5on 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 neglec5ng fric5on? Power is constant, so
, a constant.
Integra5ng with respect to 5me, and no5ng that the ini5al velocity is zero, one gets . So geqng to twice the speed takes 4 5mes as long, and the 5me to reach 60 mph is 4×1.5 = 6s. 2/20/11
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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 propor5onal to the square of the plane’s speed.) A.  by half B.  unchanged C.  doubled E.  8 9mes 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
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Recall: Center of Mass Center‐Of‐Mass Work Defini5on of the center of mass: 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
Because
where
,
is the translational kinetic energy
Net work done on collection
of objects
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=
change in translational kinetic
energy of system
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