Physics 1501: Lecture 11, Pg 1
Physics 1501: Lecture 11Physics 1501: Lecture 11 Announcements
HW 04 will be due the following Friday (same week).
Midterm 1: Monday Oct. 3Wednesday
» Replacement (Sandipan Banerjee)
» Office hours Friday instead of Wednesday
TopicsWork & EnergyScalar Product
Physics 1501: Lecture 11, Pg 2
Chap.6: Work & EnergyChap.6: Work & Energy
One of the most important concepts in physics.Alternative approach to mechanics.
Many applications beyond mechanics.Thermodynamics (movement of heat).Quantum mechanics...
Very useful tools.You will learn new (sometimes much easier) ways to
solve problems.
Physics 1501: Lecture 11, Pg 3
Definition of Work:Definition of Work:
Ingredients: Ingredients: Force ( FF ), displacement ( rr )
Work, W, of a constant force FF
acting through a displacement rr
is:
W = FF · rr = F rr cos = Fr rr
FF
rr
displace
ment
Fr
“Dot Product”
Physics 1501: Lecture 11, Pg 4
Review: Scalar Product ( or Dot Product)Review: Scalar Product ( or Dot Product)
Definition:a a · bb = ab cos
= a[b cos ] = aba
= b[a cos ] = bab
Some properties:a a ·bb = b b ·aaq(a a ·bb) = (qbb) · a a = bb · (qaa) (q is a scalar)aa · (b b + cc) = (a a ·bb) + (aa ·cc) (cc is a vector)
The dot product of perpendicular vectors is 0 !!
aa
ab bb
aa
bb
ba
Physics 1501: Lecture 11, Pg 5
Review: Examples of dot productsReview: Examples of dot products
Suppose Then
aa = 1 i i + 2 j j + 3 k k
bb = 4 i i - 5 j j + 6 k k
aa · bb = 1x4 + 2x(-5) + 3x6 = 12
aa · aa = 1x1 + 2x2 + 3x3 = 14
bb · bb = 4x4 + (-5)x(-5) + 6x6 = 77
i i · ii = j j · j j = k k · k k = 1
i i · jj = j j · k k = k k · i i = 0x
y
z
ii
jj
kk
Physics 1501: Lecture 11, Pg 6
Review: Properties of dot productsReview: Properties of dot products
Magnitude:a2 = |a|2 = a · a
= (ax i i + ay j j ) · (ax i i + ay j j )= ax
2( i i · i i ) + ay 2( j j · j j ) + 2ax ay ( i i · j j )
= ax 2 + ay
2
Pythagorian Theorem !!
aa
ax
ay
ii
j j
Physics 1501: Lecture 11, Pg 7
Review: Properties of dot productsReview: Properties of dot products
Components:aa = ax i i + ay j j + az k k = (ax , ay , az ) = (aa · i i , aa · j j , aa · k k )
Derivatives:
Apply to velocity
So if v is constant (like for UCM):
Physics 1501: Lecture 11, Pg 8
Back to the definition of Work:Back to the definition of Work:
Work, W, of a force FF acting
through a displacement rr is:
W = FF · rrFF
rr
Physics 1501: Lecture 11, Pg 9
Lecture 11, Lecture 11, ACT 1ACT 1WorkWork
A box is pulled up a rough ( > 0) incline by a rope-pulley-weight arrangement as shown below.
How many forces are doing work on the box ?
(a)(a) 2
(b)(b) 3
(c)(c) 4
Physics 1501: Lecture 11, Pg 10
Work: 1-D Example Work: 1-D Example (constant force)(constant force)
A force FF = 10N pushes a box across a frictionless floor for a distance x x = 5m.
xx
FF
Work done byby F F onon box : WF = FF ·xx = F x (since FF is parallel to xx)
WF = (10 N)x(5m) = 50 N-m.
Physics 1501: Lecture 11, Pg 11
Units:Units:
N-m (Joule) Dyne-cm (erg)
= 10-7 J
BTU = 1054 J
calorie = 4.184 J
foot-lb = 1.356 J
eV = 1.6x10-19 J
cgs othermks
Force x Distance = Work
Newton x
[M][L] / [T]2
Meter = Joule
[L] [M][L]2 / [T]2
Physics 1501: Lecture 11, Pg 12
Work and Varying ForcesWork and Varying Forces
Consider a varying force,
W = Fxx
As x 0, x dx
Fx
xx
Area = Fxx
Physics 1501: Lecture 11, Pg 13
SpringsSprings
A very common problem with a variable force is a spring.
In this spring, the force gets greater as the spring is further compressed.
Hook’s Law,
FS = - k x
x is the amount the spring is stretched or compressed from it resting position.
Fx
Active Figure
Physics 1501: Lecture 11, Pg 14
Lecture 11,Lecture 11, ACT 2ACT 2HookHook’’s Laws Law
Remember Hook’s Law,
Fx = -k x
What are the units for the constant k ?
A) B) C) D)
Physics 1501: Lecture 11, Pg 15
Lecture 11,Lecture 11, ACT 3ACT 3HookHook’’s Laws Law
0.2 kg
9 cm
8 cm
What is k for this spring ??
A) 50 N/m B) 100 N/m C) 200 N/m D) 400 N/m
Physics 1501: Lecture 11, Pg 16
What is the Work done by the Spring...What is the Work done by the Spring... The work done by the spring Ws during a displacement
from x1 to x2 is the area under the F(x) vs x plot between x1 and x2.
Ws = - 1/2 [ ( kx2) (x2) - (kx1) (x1) ]
x2 x1
F(x)
x
Ws
kx1
kx2
-kx
Ws
Active Figure
Physics 1501: Lecture 11, Pg 17
Work & Kinetic Energy:Work & Kinetic Energy:
A force FF = 10N pushes a box across a frictionlessfloor for a distance x x = 5m. The speed of the box is v1 before the push, and v2 after the push.
xx
FFv1 v2
ii
m
Physics 1501: Lecture 11, Pg 18
Work & Kinetic Energy...Work & Kinetic Energy...
Since the force FF is constant, acceleration aa will be constant. We have shown that for constant a:
W = (F) · d = ma · dFor constant a, a = (v-v0)/t
also, d = vt = (1/2) (v+v0)t
xx
FFv1 v2
aa
ii
m
Physics 1501: Lecture 11, Pg 19
Work & Kinetic Energy...Work & Kinetic Energy...
Altogether,
W = (F) · d = ma ·d = m ((v-v0)/t) · 1/2 (v+v0)t
W = (1/2) m (v2 - v02)
Define Kinetic Energy K: K = 1/2mv2
K2 - K1 = WF
WF = K (Work kinetic-energy theorem)(Work kinetic-energy theorem)
xx
FFv1 v2
aa
ii
m
Physics 1501: Lecture 11, Pg 20
Work Kinetic-Energy Theorem:Work Kinetic-Energy Theorem:
{NetNet WorkWork done on object}=
{changechange in kinetic energy kinetic energy of object}
We’ll prove this for a variable force later.
Physics 1501: Lecture 11, Pg 21
xx
vo
m
to
F
ExampleExampleWork Kinetic-Energy TheoremWork Kinetic-Energy Theorem
Km = 1/2 m v2 - 1/2 m vo2
= 0 - 1/2 m vo2
Km = Wspring
Wspring = -1/2 k x2 - xo2 )
= - 1/2 k x2
(assuming x0=0)so : 1/2 m vo
2 = 1/2 k x2
and : x = ( m vo
2 /k ) 1/2
How much will the spring compress to bring the object to a stop if the object is moving initially at a constant velocity (vo) on frictionless surface as shown below ?
spring compressed
spring at an equilibrium position
V=0
t m
Physics 1501: Lecture 11, Pg 22
Lecture 11, Lecture 11, ACT 4ACT 4Kinetic EnergyKinetic 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 ?
(a)(a) 1/41/4 (b)(b) 1/2 (c)(c) 1 (d) 2 (e) 4
Physics 1501: Lecture 11, Pg 23
Lecture 11, Lecture 11, ACT 5ACT 5Kinetic EnergyKinetic Energy
To practice your pitching you use two baseballs. The first time you use a softball: you throw a slow curve and clock the speed at 50 mph (~25 m/s). The second time you use a baseball (about half the mass of a softball) and go with high heat: 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)(a) 1/4 (b)(b) 1/2 (c)(c) 1 (d) 2 (e) 4
Physics 1501: Lecture 11, Pg 24
Lecture 11, Lecture 11, ACT 6ACT 6Work & EnergyWork & Energy
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. > 0) which slows them down to a stop.Which one will go farther before stopping ?
(a)(a) m1 (b)(b) m2 (c)(c) they will go the same distance
m1
m2
Physics 1501: Lecture 11, Pg 25
Lecture 11, Lecture 11, ACT 7ACT 7Work & EnergyWork & Energy
You like to drive home fast, slam on your brakes at the bottom 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 mama. 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)(a) The same distance. Not so exciting.The same distance. Not so exciting.
b) 2 times as far (only 7/10 of the way up the driveway)
(c)(c) twice as far, right to the door. Whoopee!twice as far, right to the door. Whoopee!
(d)(d) four times as far. Crashes into house. Sorry Ma.four times as far. Crashes into house. Sorry Ma.
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