Transcript Physics 131: Lecture 14 Notes

Physics 151: Lecture 12

Topics :
(Text Ch. 7.1-7.4)
» Work & Energy.
» Work of a constant force.
» Work of a non-constant force.
» Work - Energy Theorem.
Physics 151: Lecture 12, Pg 1
Forms of Energy

Kinetic: Energy of motion.
A car on the highway has kinetic energy.
We have to remove this energy to stop it.
The breaks of a car get HOT !
This is an example of turning one form of energy into
another (thermal energy).
Physics 151: Lecture 12, Pg 2
See text: 7-1
Energy Conservation

Energy cannot be destroyed or created.
Just changed from one form to another.

We say energy is conserved !
Animation
True for any isolated system.
i.e when we put on the brakes, the kinetic energy of the car
is turned into heat using friction in the brakes. The total
energy of the “car-breaks-road-atmosphere” system is the
same.
The energy of the car “alone” is not conserved...
» It is reduced by the braking.

Doing “work” on an otherwise isolated system will change it’s
“energy”...
Physics 151: Lecture 12, Pg 3
See text: 7-1
Definition of Work:
Ingredients: Force ( F ), displacement (  r )
Work, W, of a constant force F
acting through a displacement  r
is:
F

Fr
r
.
W = F  r = F  r cos  = Fr  r
“Dot Product”
Physics 151: Lecture 12, Pg 4
Definition of Work...

Only the component of F along the displacement is doing
work.
Example: Train on a track.
F

r
F cos 
Physics 151: Lecture 12, Pg 5
See text: 7.2
Review: Scalar Product ( or Dot Product)
a
Definition:
ba
.
a b = ab cos 

= a[b cos ] = aba
b
a
= b[a cos ] = bab

Some properties:
a b= b a
q(a b) = (qb) a = b (qa)
a (b + c) = (a b) + (a c)
.
.
.
.
.
.
.
.
b
ab
(q is a scalar)
(c is a vector)
The dot product of perpendicular vectors is 0 !!
Physics 151: Lecture 12, Pg 6
See text: 7.2
Review: Examples of dot products
y
.i=j.j=k.k=1
i.j=j.k=k.i=0
i
j
z
Suppose
a=1i+2j+3k
b=4i -5j+6k
k
i
x
Then
.
.
.
a b = 1x4 + 2x(-5) + 3x6 = 12
a a = 1x1 +
2x2 + 3x3 = 14
b b = 4x4 + (-5)x(-5) + 6x6 = 77
Physics 151: Lecture 12, Pg 7
See text: 7.2
Review: Properties of dot products

Magnitude:
a2 = |a|2
.
=a a
= (ax i + ay j ) (ax i + ay j )
= ax 2( i i ) + ay 2( j j ) + 2ax ay ( i
= ax 2 + ay 2
.
Pythagorian Theorem !!
.
.
a
.j)
ay
ax
j
i
Physics 151: Lecture 12, Pg 8
Review: Properties of dot products

Components:
a = ax i + ay j + az k = (ax , ay , az ) = (a

Derivatives:
.i,a.j,a.k
)
d
da
db
(ab ) 
b  a
dt
dt
dt
Apply to velocity
d 2 d
dv
dv
v  (v v ) 
v  v 
 2v  a
dt
dt
dt
dt
So if v is constant (like for UCM):
d 2
v  2v  a  0
dt
Physics 151: Lecture 12, Pg 9
Lecture 12, ACT 1
Work

A box is pulled up a rough (m > 0) incline by a rope-pulleyweight arrangement as shown below.
How many forces are doing work on the box ?
(a) 2
(b) 3
(c) 4
Is the work done by F
positive or negative?
Physics 151: Lecture 12, Pg 10
See text: 7-1
Work: 1-D Example
(constant force)

A force F = 10N pushes a box across a frictionless
floor for a distance x = 5m.
F
x
Work done by F on box :
WF = F . x = F x
(since F is parallel to x)
WF = (10 N)x(5m) = 50 N-m.
See example 7-1: Pushing a trunk.
Physics 151: Lecture 12, Pg 11
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 151: Lecture 12, Pg 12
Text : 7.3
Work and Varying Forces

Consider a varying force,
Area = Fxx
Fx
x

W = Fxx

As x  0, x  dx
x
dW  Fx dx
W   Fx dx
Physics 151: Lecture 12, Pg 13
Text : 7.3
Springs

A very common problem with a variable force is a spring.
x
F

In this spring, the force gets greater as the spring is further
compressed.

Hook’s Law,
FS = - k x
Animation
x is the amount the spring is stretched or compressed
from it resting position.
Physics 151: Lecture 12, Pg 14
Lecture 12, ACT 2
Hook’s Law

Remember Hook’s Law,
Fx = -k x
What are the units for the constant k ?
A) kg  m
s2
2
B) kg  m
s2
C)
kg
s2
D) kg  m
2
s2
Physics 151: Lecture 12, Pg 15
Lecture 12, ACT 3
Hook’s Law
8 cm
9 cm
What is k for this spring ??
0.2 kg
A) 50 N/m
B) 100 N/m C) 200 N/m D) 400 N/m
Physics 151: Lecture 12, Pg 16
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.
x2
Ws   F ( x) dx
F(x)
x1
x2
x1
x2
  (  kx) dx
x1
x
1
2
x
  kx
x
2
Ws
1
Ws   k x22  x12 
2
-kx
2
kx1
1
kx2
Ws = - 1/2 [ ( kx2) (x2) - (kx1) (x1)U]  Ws 
1
k x22  x12 
2
Physics 151: Lecture 12, Pg 17
Work & Kinetic Energy:

A force F = 10N pushes a box across a frictionless
floor for a distance x = 5m. The speed of the box is v1
before the push, and v2 after the push.
v1
v2
F
m
i
x
Physics 151: Lecture 12, Pg 18
Work & Kinetic Energy...

Since the force F is constant, acceleration a will be
constant. We have shown that for constant a:
S W = (S F).d = ma.d
For constant a, a = (v-v0)/t
also, d = vavt = 1/2 (v+v0)t
v1
v2
F
m
a
i
x
Physics 151: Lecture 12, Pg 19
Work & Kinetic Energy...

Altogether,
.
.
S W = (S F).d = ma d = m [(v-v0)/t) (1/2 (v+v0)t]
S W = 1/2 m ( v2 - v02 ) = (1/2 m v2 ) - (1/2 m v02 )

K = 1/2mv2
Define Kinetic Energy K:
K2 - K1 = WF
WF = K (Work kinetic-energy theorem)
v1
v2
F
m
a
i
x
Physics 151: Lecture 12, Pg 20
See text: 7-4
Work Kinetic-Energy Theorem:
Net Work done on object = change in kinetic energy of object
Wnet  K
 K 2  K1



1
1
2
2
mv 2  mv1
2
2
Is this applicable also for a variable force ?
YES
Physics 151: Lecture 12, Pg 21
Lecture 12, ACT 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 ?
(a) 1/4
(b) 1/2
(c) 1
(d) 2
(e) 4
Physics 151: Lecture 12, Pg 22
Example
Work Kinetic-Energy Theorem

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 ?
to vo
F
m
spring at an equilibrium position
x = ( m vo2 / k ) 1/2
x
V=0
t
m
spring compressed
Physics 151: Lecture 12, Pg 23
Act 4

Two clowns are launched from the same spring-loaded
circus cannon with the spring compressed the same
distance each time. Clown A has a 40-kg mass; clown B a
60-kg mass. The relation between their kinetic energies at
the instant of launch is :
a) KA = 4 KB
b) KA = 2 KB
c) KA = KB
d) KB = 2 KA
e) KB = 4 KA
Physics 151: Lecture 12, Pg 24
Act 4.b

Two clowns are launched from the same spring-loaded
circus cannon with the spring compressed the same
distance each time. Clown A has a 40-kg mass; clown B a
60-kg mass. The relation between their speeds at the
instant of launch is:
a) vA = 3/2 vB
b) vA = (3/2 )1/2 vB
c) vA = vB
d) vB = (3/2 )1/2 vA
e) vB = 3/2 vA
Physics 151: Lecture 12, Pg 25
Recap of today’s lecture

Work & Energy. (Text: 7.1-4)
Discussion.
Definition.
Work of a constant and non-constant forces.

Work - Energy Theorem.

Physics 151: Lecture 12, Pg 26