Transcript sy10_oct09

Physics 207, Lecture 10, Oct. 9

•
•
MidTerm I
Exams will be returned in your next discussion section
Regrades: Write down, on a separate sheet, what you
want regraded and why.
Mean: 64.6
Median: 67
Std. Dev.: 19.0
Range: High 100 Low 5
Solution posted on
http://my.wisc.edu
Nominal curve (conservative):
86-100 A
70-85 B or A/B
40-69 C or B/C
35-40 marginal
25-35 D
Below 25 F
Physics 207: Lecture 10, Pg 1
Physics 207, Lecture 10, Oct. 9
Agenda: Chapter 7, Work and Energy Transfer
• Definition of Work (a scalar quantity)
• Variable force devices (e.g., Hooke’s Law spring)
• Work/Energy Theorem
•
•
 W = DK
Kinetic Energy
 K = 1/2 mv2
Power
 P = dW / dt = F · v
Assignment: For Wednesday read Chapter 8
 WebAssign Problem Set 4 due Tuesday next week (start now)
Physics 207: Lecture 10, Pg 2
Work & Energy

One of the most important concepts in physics.
 Alternative approach to mechanics.

Many applications beyond mechanics.
 Thermodynamics (movement of heat or particles).
 Quantum mechanics...

Very useful tools.
 You will learn a complementary approach (often
much easier) way to solve problems. But there is
no free lunch….easier but there are fewer details
that are explicitly known.
Physics 207: Lecture 10, Pg 3
See text: 7-1
Energy Conservation

Energy cannot be destroyed or created.
Just changed from one form to another.

We say energy is conserved !
True for any isolated system.

Doing “work” on an otherwise isolated system will
change it’s “energy”...
Physics 207: Lecture 10, Pg 4
See text: 7-1
Definition of Work, The basics
Ingredients: Force ( F ), displacement ( D r )
Work, W, of a constant force F
acting through a displacement D r
is:
W = F ·D r
F

Dr
(Work is a scalar)
“Scalar or Dot Product”
Work tells you something about what happened on the path!
Did something do work on you? Did you do work on something?
Simplest case (no frictional forces and no non-contact forces)
Did your speed change? ( what happened to | v | !!!)
Physics 207: Lecture 10, Pg 5
Remember that a path evolves with time
and acceleration implies a force acting on an object
v
path
and time
a
t
a = atang + aradial
a
a
a = a + a
a =0
F = Ftang + Fradial
Two possible options:

Change in the magnitude of v
a
=0
Change in the direction of v
a
=0
A tangetial force is the important one for work!
 How long (time dependence) gives the kinematics
 The distance over which this forceTang is applied: Work
Physics 207: Lecture 10, Pg 6
Definition of Work...
Only the component of F along the path (i.e.
“displacement”) does work.
The vector dot product does that automatically.
 Example: Train on a track.

F

Dr
F cos 
If we know
the angle the force makes
with the track, the dot product
gives us F cos  and Dr
Physics 207: Lecture 10, Pg 7
Review: Scalar Product (or Dot Product) 7.3

Useful for performing projections.
A

A  î = Ax
î

Ay
Ax
Calculation is simple in terms of components.
A  B = (Ax )(Bx) + (Ay )(By ) + (Az )(Bz )
Calculation also in terms of magnitudes and relative angles.
A  B ≡ | A | | B | cos 
You choose the way that works best for you!
Physics 207: Lecture 10, Pg 8
See text: 7-1
Work: 1-D Example
(constant force)
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A force F = 10 N pushes a box across a frictionless
floor for a distance Dx = 5 m.
Finish
Start
F
 = 0°
Dx
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

Work is A  B ≡ | A | | B | cos  = F Dx = 10 x 5 N m = 50 J
1 Nm is defined to be 1 Joule and this is a unit of energy
Work reflects energy transfer
See example 7-1: Pushing a trunk.
Physics 207: Lecture 10, Pg 9
See text: 7-1
Units:
Force x Distance = Work
Newton x Meter = Joule
[M][L] / [T]2 [L]
[M][L]2 / [T]2
mks
N-m (Joule)
cgs
Dyne-cm (erg)
= 10-7 J
Other
BTU
calorie
foot-lb
eV
= 1054 J
= 4.184 J
= 1.356 J
= 1.6x10-19 J
Physics 207: Lecture 10, Pg 10
See text: 7-1
Work: 1-D 2nd Example
(constant force)

A force F = 10 N pushes a box across a frictionless
floor for a distance Dx = 5 m.
Finish
Start
 = 180°
F
Dx

Work is A  B ≡ | A | | B | cos  = F Dx (-1) = -10 x 5 N m = -50 J

Work reflects energy transfer
See example 7-1: Pushing a trunk.
Physics 207: Lecture 10, Pg 11
See text: 7-1
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Work: 1-D 3rd Example
(constant force)
A force F = 10 N pushes a box across a frictionless
floor for a distance Dx = 5 m.
Start
F
Finish
 = -45°
Dx

Work is A  B ≡ | A | | B | cos  = F Dx 0.71= 50 x 0.71 Nm = 35 J
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Work reflects energy transfer
See example 7-1: Pushing a trunk.
Physics 207: Lecture 10, Pg 12
Text : 7.3
Work and Varying Forces

Area = Fx Dx
F is increasing
Here W = F ·D r
becomes dW = F dx
xf
Consider a varying force F(x)
Fx
W
x
Dx
xi
Finish
Start
F
  F ( x ) dx
F
 = 0°
Dx
Work is a scalar, the catch is that there is no time/position info on hand
Physics 207: Lecture 10, Pg 13
Lecture 10, Exercise 1
Work in the presence of friction and non-contact forces
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A box is pulled up a rough (m > 0) incline by a ropepulley-weight arrangement as shown below.
 How many forces are doing work on the box ?
 Of these which are positive and which are negative?
 Use a Force Body Diagram
(A) 2
 Compare force and path
(B) 3
v
(C) 4
Physics 207: Lecture 10, Pg 14
Lecture 10, Exercise 1
Work in the presence of friction and non-contact forces

A box is pulled up a rough (m > 0) incline by a ropepulley-weight arrangement as shown below.
 How many forces are doing work on the box ?
 And which are positive and which are negative?
 Use a Force Body Diagram
(A) 2
v
T
(B) 3 is correct
N
(C) 4
mg
f
Physics 207: Lecture 10, Pg 15
Text : 7.3
A variable force device: A Hooke’s Law Spring

Springs are everywhere, (probe microscopes, DNA, an
effective interaction between atoms)
Rest or equilibrium position
Dx


F(work done on spring)
In this spring, the magnitude of the force increases as
the spring is further compressed (a displacement).
Hooke’s Law,
Active Figure
FS = - k Dx
Dx is the amount the spring is stretched or compressed
from it resting position.
Physics 207: Lecture 10, Pg 16
Lecture 10, Exercise 2
Hooke’s Law
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Remember Hooke’s Law,
Fx = -k Dx
What are the units for the constant k ?
(A)
kg m 2
s2
(B)
kg m
s2
(C)
kg
s2
(D)
kg 2 m
s2
F is in kg m/s2 and dividing by m gives kg/s2 or N/m
Physics 207: Lecture 10, Pg 17
Lecture 10, Exercise 3
Hooke’s Law
8 cm
9 cm
What is the spring constant “k” ?
0.50 kg
(A) 50 N/m
(B) 100 N/m (C) 400 N/m (D) 500 N/m
Physics 207: Lecture 10, Pg 18
Lecture 10, Exercise 3
Hooke’s Law
8 cm
9 cm
What is the spring constant “k” ?
Fspring
0.50 kg
(A) 50 N/m
SF = 0 = Fs – mg = k Dx - mg
Use k = mg/Dx = 5 N / 0.01 m
(B) 100 N/m (C) 400 N/m (D) 500 N/m
mg
Physics 207: Lecture 10, Pg 19
Force (N)
F-x relation for a foot arch:
Displacement (mm)
Physics 207: Lecture 10, Pg 20
F-x relation for a single DNA molecule
Physics 207: Lecture 10, Pg 21
Measurement technique: optical tweezers
Physics 207: Lecture 10, Pg 22
Work & Kinetic Energy:
Energy transfer involving changes in speed
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A force, F = 10 N, pushes a box across a frictionless
floor for a distance Dx = 5m.
The speed of the box is v1 before the push, and v2
after the push.
Consider only this force and the box
Relate the work to the kinetic energy of the box
F
v
v1
F’
2
m
i
Dx
Physics 207: Lecture 10, Pg 23
See text: 7-4
Work Kinetic-Energy Theorem:
{Net Work done on object}
=
{change in kinetic energy of object}
Wnet  DK
 K 2  K1
(final – initial)
1
1
2
2
 mv 2  mv1
2
2
Physics 207: Lecture 10, Pg 24
Example: Work Kinetic-Energy Theorem
•
How much will the spring compress (i.e. Dx) to bring
the object to a stop (i.e., v = 0 ) if the object is moving
initially at a constant velocity (vo) on frictionless surface
as shown below ?
to vo
Notice that the
spring force is
opposite to the
displacemant.
F
m
spring at an equilibrium position
For the mass m,
work is negative
Dx
V=0
t
m
For the spring, work
is positive
spring compressed
Physics 207: Lecture 10, Pg 25
Example: Work Kinetic-Energy Theorem
•
How much will the spring compress (i.e. Dx = xf - xi)
to bring the object to a stop (i.e., v = 0 ) if the object is
moving initially at a constant velocity (vo) on frictionless
surface as shown below ?
xf
to vo
Wbox
F
xi
xf
m
spring at an equilibrium position
Dx
V=0
t
  F ( x ) dx
Wbox
xi
x
f
Wbox  - kx |
xi
1
2
Wbox
m
spring compressed
   kx dx
2
 - k Dx  DK
1
2
2
- k Dx  m0  mv
1
2
2
1
2
2
1
2
2
0
Physics 207: Lecture 10, Pg 26
Lecture 10, Exercise 4
Kinetic Energy
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To practice your pitching you use two baseballs.
The first time you throw a slow curve and clock the
speed at 50 mph (~25 m/s). The second time you
go with high heat and the radar gun clocks the pitch
at 100 mph. What is the ratio of the kinetic energy
of the fast ball versus the curve ball ?
(A) 1/4
(B) 1/2
(C) 1
(D) 2
(E) 4
Physics 207: Lecture 10, Pg 27
Lecture 10, Exercise 4
Kinetic Energy

To practice your pitching you use two baseballs.
The first time you throw a slow curve and clock the
speed at 50 mph (~25 m/s). The second time you
go with high heat and the radar gun clocks the pitch
at 100 mph. What is the ratio of the kinetic energy
of the fast ball versus the curve ball ?
KE2/KE1 = ½ mv22 / ½ mv12 = 1002 / 502 = 4
(A) 1/4
(B) 1/2
(C) 1
(D) 2
(E) 4
Physics 207: Lecture 10, Pg 28
Lecture 10, Exercise 5
Work & Friction
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

Two blocks having mass 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
(i.e. m > 0) which slows them down to a stop.
Which one will go farther before stopping?
Hint: How much work does friction do on each block ?
(A) m1
(B) m2
m1
m2
(C) They will go the same distance
v1
v2
Physics 207: Lecture 10, Pg 29
Lecture 10, Exercise 5
Work & Friction
 W = F d = - m N d = - m mg d = DK = 0 – ½ mv2

- m m1g d1 = - m m2g d2  d1 / d2 = m2 / m1
(A) m1
(B) m2
m1
m2
(C) They will go the same distance
v1
v2
Physics 207: Lecture 10, Pg 30
Lecture 10, Exercise 6
Work & Friction

You like to drive home fast, slam on your brakes at the start of
the driveway, and screech to a stop “laying rubber” all the
way. It’s particularly fun when your mother is in the car with
you. You practice this trick driving at 20 mph and with some
groceries in your car with the same mass as your mother. You
find that you only travel half way up the driveway. Thus when
your mom joins you in the car, you try it driving twice as fast.
How far will you go this time ?
A. The same distance. Not so exciting.
B.  2 times as far (only ~7/10 of the way up the driveway)
C. Twice as far, right to the door. Whoopee!
D. Four times as far crashing into the house. (Oops.)
Physics 207: Lecture 10, Pg 31
Lecture 10, Exercise 6
Work & Friction
W = F d = - m N d = - m mg d = DK = 0 – ½ mv2
W1= - m mg d1= DK1= 0 – ½ mv12
W2= - m mg d2= DK2= 0 – ½ m(2v1)2 = – 4 (½ mv12)

- m mg d2= – 4 (m mg d1)  d2= – 4 d1
A. The same distance. Not so exciting.
B.  2 times as far (only ~7/10 of the way up the driveway)
C. Twice as far, right to the door. Whoopee!
D. Four times as far crashing into the house. (Oops.)
Physics 207: Lecture 10, Pg 32
Work & Power:





Two cars go up a hill, a Corvette and a ordinary
Chevy Malibu. Both cars have the same mass.
Assuming identical friction, both engines do the same
amount of work to get up the hill.
Are the cars essentially the same ?
NO. The Corvette gets up the hill quicker
It has a more powerful engine.
Physics 207: Lecture 10, Pg 33
Work & Power:



Power is the rate at which work is done.
W
Average Power is,
P 
Instantaneous Power is,
Dt
dW
P
dt
Physics 207: Lecture 10, Pg 34
Lecture 10, Exercise 8
Work & Power

Starting from rest, a car drives up a hill at constant
acceleration and then suddenly stops at the top. The
instantaneous power delivered by the engine during
this drive looks like which of the following,

(A)


(B)
(C)
time
time
time
Physics 207: Lecture 10, Pg 35
Lecture 10, Exercise 8
Work & Power

P = dW / dt and W = F d = (m mg cos   mg sin ) d
and d = ½ a t2 (constant accelation)
So W = F ½ a t2  P = F a t = F v

(A)
time

(B)
time

(C)
time
Physics 207: Lecture 10, Pg 36
Work & Power:

Power is the rate at which work is done.
Average
Power:
W
P
Dt



Instantaneous
Power:
Units (SI) are
Watts (W):
dW
1 W = 1 J / 1s
P
dt
Simple Example 1 :
A person of mass 80.0 kg walks up to 3rd floor
(12.0m). If he/she climbs in 20.0 sec what is the
average power used.
Pavg = F h / t = mgh / t = 80.0 x 9.80 x 12.0 / 20.0 W
P = 470. W
Physics 207: Lecture 10, Pg 37
Work & Power:
Simple Example 2 :
Engine of a jet develops a trust of 15,000 N when plane is
flying at 300 m/s. What is the horsepower of the engine ?
dW
P
dt


W  F  Dx
 
dW  F  dx

dW  dx
P
F
dt
dt
 
P  F v
P=Fv
P = (15,000 N) (300 m/s) = 4.5 x 106 W
= (4.5 x 106 W) (1 hp / 746 W) ~ 6,000 hp !
Physics 207: Lecture 10, Pg 38
Physics 207, Lecture 10, Recap
Agenda: Chapter 7, Work and Energy Transfer
• Definition of Work (a scalar quantity)
• Variable force devices (e.g., Hooke’s Law spring)
• Work/Energy Theorem
•
•
 W = DK
Kinetic Energy
 K = 1/2 mv2
Power
 P = dW / dt = F · v
Assignment: For Wednesday read Chapter 8
 WebAssign Problem Set 4 due Tuesday next week (start now)
Physics 207: Lecture 10, Pg 39