1. Chapter 6 Work and Energy
Work (dot product)
Work by gravity, by spring
Kinetic energy, power
Work-kinetic energy theorem
C.M. system
4/22/2017
Phys 201, Spring 2011

2. Definition of Work: Constant Force
Ingredients: Force (F), displacement (Δr)
Work, W, of a constant force F
acting through a displacement Δr :
Vector
“Dot Product”
Work is a scalar.
Units are Newton meter: Joule = N x m
4/22/2017
Phys 201, Spring 2011

3. Work and Force Direction
Work is done in lifting the box (you do +work; gravity -)
No work is done on the bucket when held still,
or to move horizontally
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Phys 201, Spring 2011

4. More on “dot product” (or scalar product)θ a b ba
Definition:
a.b = ab cos θ
= a[b cos θ] = aba
= b[a cos θ] = bab
Some properties:
a . b = b . a
q(a . b) = (qb) . a = b .(qa) (q is a scalar)
a .(b + c) = (a . b) + (a . c) (c is a vector)
The dot product of perpendicular vectors is 0 !! θ a ab b
4/22/2017
Phys 201, Spring 2011

5. Examples of dot productsy z i j k
i . i = j . j = k . k = 1
i . j = j . k = k . i = 0
Suppose
Then
a = 1 i + 2 j + 3 k
b = 4 i - 5 j + 6 k
a . b = 1x x(-5) + 3x6 = 12
a . a = 1x x x3 = 14
b . b = 4x4 + (-5)x(-5) + 6x6 = 77
4/22/2017
Phys 201, Spring 2011

6. More properties of dot products
Components:
a = ax i + ay j + az k = (ax , ay , az) = (a . i, a . j, a . k)
Derivatives:
Apply to velocity
So if v is constant
(like for uniform circular motion):
4/22/2017
Phys 201, Spring 2011

7. Which of the statements below is correct?
The scalar product of two vectors can be negative.
AcB=c(BA), where c is a constant.
The scalar product can be non-zero even if two of the three components of the two vectors are equal to zero.
If A = B+C, then DA = DB+DC
All of the above statements are correct.
4/22/2017
Phys 201, Spring 2011

8. What about multiple forces?
Suppose FNET = F1 + F2 and the
displacement is Δr.
The work done by each force is:
W1 = F1 . Δr W2 = F2 . Δr
WTOT = W1 + W2
= F1 . Δr + F2 . Δr
= (F1 + F2 ) . Δr
WTOT = FTOT . Δr It’s the total force that matters!!
FNET F1 Δr F2
4/22/2017
Phys 201, Spring 2011

9. Question 1 W T FN V
You are towing a car up a hill with constant velocity. The work done on the car by the normal force is:
1. positive 2. negative 3. zero
correct
Normal force is perpendicular to displacement
cos θ = 0
4/22/2017
Phys 201, Spring 2011

10. Question 2 W T FN V
You are towing a car up a hill with constant velocity. The work done on the car by the gravitational force is:
1. positive 2. negative 3. zero
correct
There is a non-zero component of gravitational force pointing opposite the direction of motion.
4/22/2017
Phys 201, Spring 2011

11. Question 3 W T FN V
You are towing a car up a hill with constant velocity. The work done on the car by the tension force is:
1. positive 2. negative 3. zero
correct
T is pointing in the direction of motion - therefore, work done by this force is positive.
4/22/2017
Phys 201, Spring 2011

12. Question 4 W T FN V
You are towing a car up a hill with constant velocity. The total work done on the car by all forces is:
1. positive 2. negative 3. zero
correct
Constant velocity implies that there is no net force acting on the car, so there is no work being done overall
4/22/2017
Phys 201, Spring 2011

13. Lifting an object:
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.
(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
(c) the net work done on the truck:
W = Wc + Wg = 0
4/22/2017
Phys 201, Spring 2011

14. Work by a variable force:
Total work, W, of a force F acting through a displacement Δx = dx î is an integration: Fx Fi x
Δxi
Work = Area under F(x) curve
4/22/2017
Phys 201, Spring 2011

15. 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:
4/22/2017
Phys 201, Spring 2011

16. A person pushes against a spring, the opposite end of which is attached to a fixed wall. The spring compresses. Is the work done by the spring on the person positive, negative or zero?
positive
negative
zero
4/22/2017
Phys 201, Spring 2011

17. It is convertible into other forms without loss (i.e it is conserved)
Energy
Energy is that quality of a substance or object
which “causes something to happen”;
or “capability of exerting forces”;
or “ability to do work”…
The vagueness of the definition is due to the fact
that energy can result in many effects.
Electrical Energy
Chemical Energy
Mechanical Energy
Nuclear Energy
The force that accompanies the “capability” takes various forms (labeling the energy).
Energy (capability) can increase or decrease.
That is, energy can be decreased by exerting a force that does work, and can be increased if an external force does work.
It is convertible into other forms without loss (i.e it is conserved)
4/22/2017
Phys 201, Spring 2011

18. Kinetic Energy The “energy of motion”. W = mV - m V W = F d K.E. = mV
Work done on the object increases its energy,
-- by how much? (i.e. how to calculate the value?)
W = F d
V = V a d 2
F = ma
a =
2 d
V - V 2
W = (ma) d
Turn off sound
2 d
V - V 2
W = m d
W = mV m V 1 2
K.E. = mV 1 2
4/22/2017
Phys 201, Spring 2011

19. Definition of Kinetic Energy
Kinetic energy (K.E.) of a particle of mass m moving with speed v is defined as K.E. = ½mv2 Kinetic energy is a useful concept because of the Work/Kinetic Energy theorem, which relates the work done on an object to the change in kinetic energy.
4/22/2017
Phys 201, Spring 2011

20. A falling object
What is the speed of an object that starts at rest and then falls a vertical distance H? Work done by gravitational force WG = FΔr = mgH Work/Kinetic Energy Theorem:
v0 = 0 mg j r H v
4/22/2017
Phys 201, Spring 2011

21. A skier of mass 50 kg is moving at speed 10 m/s at point P1 down a ski slope with negligible friction. What is the skier’s kinetic energy when she is at point P2, 20 m below P1?
2500 J
9800 J
12300 J
13100 J
15000 J
Work-kinetic energy theorem:
KEf – KEi = Wg = mgH
KEf = ½×50× ×9.8×20 = 2500 J J
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Phys 201, Spring 2011

22. 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.
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Phys 201, Spring 2011

23. Problem: Work and Energy
Two blocks have masses m1 and m2, where m1 > m2. They are sliding on a frictionless floor and have the same kinetic 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
Solution next page  m1 m2
4/22/2017
Phys 201, Spring 2011

24. Problem: Work and Energy (Solution)
The net work done to stop a box is -fD = -μmgD. The work-kinetic energy theorem says that for any object, WNET = ΔK so WNET=Kf-Ki=0-Ki. Since the boxes start out with the same kinetic energy, we have μm1gD1=μm2gD2 and D1/D2=m2/m1. Since m1>m2, we must have D2>D1. m2 D2 m1 D1
4/22/2017
Phys 201, Spring 2011

25. Problem:
The magnitude of the single force acting on a particle of mass m is given by F = bx2, where b is a constant. The particle starts from rest at x=0. After it travels a distance L, determine its (a) kinetic energy and (b) speed.
work done by force:
So (a) kinetic energy is K=bL3/3
(b) Since ½mv2=bL3/3,
4/22/2017
Phys 201, Spring 2011

26. Power: P: Work done by unit time. Units: Watt 4/22/2017
Phys 201, Spring 2011

27. Question
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 neglecting friction?
Power is constant, so , a constant.
Integrating with respect to time, and noting that the initial velocity is zero, one gets So getting to twice the speed takes 4 times as long, and the time to reach 60 mph is 4×1.5 = 6s.
4/22/2017
Phys 201, Spring 2011

28. 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 proportional to the square of the plane’s speed.)
by half
unchanged
doubled
quadrupled
8 times
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
4/22/2017
Phys 201, Spring 2011

29. Recall: Center of Mass Definition of the center of mass: Because ,
4/22/2017
Phys 201, Spring 2011

30. Net work done on collection
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 =
4/22/2017
Phys 201, Spring 2011