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Physics 111: Lecture 9
Today’s Agenda





Work & Energy
Discussion
Definition
Dot Product
Work of a constant force
Work/kinetic energy theorem
Work of multiple constant forces
Comments
Physics 111: Lecture 9, Pg 1
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 111: Lecture 9, Pg 2
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 111: Lecture 9, Pg 3
Mass = Energy (but not in Physics 111)

Particle Physics:
E = 1010 eV
(a)
e+
e+ 5,000,000,000 V
(b)
(c)
- 5,000,000,000 V
M
E = MC2
( poof ! )
Physics 111: Lecture 9, Pg 4
Wilberforce
Energy Conservation
Returning
Can

Energy cannot be destroyed or created.
Just changed from one form to another.

We say energy is conserved!
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 isolated system will change its “energy”...
Physics 111: Lecture 9, Pg 5
Definition of Work:
Ingredients: Force (F), displacement (r)
Work, W, of a constant force F
acting through a displacement r
is:
W = F r = F r cos  = Fr r
F

Fr
r
“Dot Product”
Physics 111: Lecture 9, Pg 6
Definition of Work...
Hairdryer

Only the component of F along the displacement is doing
work.
Example: Train on a track.
F

r
F cos 
Physics 111: Lecture 9, Pg 7
Aside: Dot Product (or Scalar Product)
a
Definition:
.
a b = ab cos 
= a[b cos ] = aba
ba

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 111: Lecture 9, Pg 8
Aside: 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 111: Lecture 9, Pg 9
Aside: 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
.
Pythagorean Theorem!!
.
.
a
. j)
ay
ax
j
i
Physics 111: Lecture 9, Pg 10
Aside: 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 111: Lecture 9, Pg 11
Back to the definition of Work:
Skateboard
Work, W, of a force F acting
through a displacement  r is:
W = F  r
F
r
Physics 111: Lecture 9, Pg 12
Lecture 9, Act 1
Work & Energy

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
Physics 111: Lecture 9, Pg 13
Lecture 9, Act 1
Solution

Draw FBD of box:

Consider direction of
motion of the box

Any force not perpendicular
to the motion will do work:
T
N
v
f
N does no work (perp. to v)
T does positive work
f does negative work
3 forces
do work
mg
mg does negative work
Physics 111: Lecture 9, Pg 14
Work: 1-D Example
(constant force)

A force F = 10 N pushes a box across a frictionless
floor for a distance x = 5 m.
F
x
Work done by F on box :
WF = Fx = F x
(since F is parallel to x)
WF = (10 N) x (5 m) = 50 Joules (J)
Physics 111: Lecture 9, Pg 15
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 111: Lecture 9, Pg 16
Work & Kinetic Energy:

A force F = 10 N pushes a box across a frictionless
floor for a distance x = 5 m. The speed of the box is v1
before the push and v2 after the push.
v1
v2
F
m
i
x
Physics 111: Lecture 9, Pg 17
Work & Kinetic Energy...

Since the force F is constant, acceleration a will be
constant. We have shown that for constant a:
v22 - v12 = 2a(x2-x1) = 2ax.
1/ mv 2 - 1/ mv 2 = max
multiply by 1/2m:
2
2
2
1
1/ mv 2 - 1/ mv 2 = Fx
But F = ma
2
2
2
1
v1
v2
F
m
a
i
x
Physics 111: Lecture 9, Pg 18
Work & Kinetic Energy...

So we find that
1/2mv22 - 1/2mv12 = Fx = WF

Define Kinetic Energy K:
K = 1/2mv2
K2 - K1 = WF
WF = K (Work/kinetic energy theorem)
v2
v1
F
m
a
i
x
Physics 111: Lecture 9, Pg 19
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
We’ll prove this for a variable force later.
Physics 111: Lecture 9, Pg 20
Lecture 9, Act 2
Work & 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 (i.e. m > 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
m1
m2
Physics 111: Lecture 9, Pg 21
Lecture 9, Act 2
Solution


The work-energy theorem says that for any object WNET = K
In this example the only force that does work is friction (since
both N and mg are perpendicular to the block’s motion).
N
f
m
mg
Physics 111: Lecture 9, Pg 22
Lecture 9, Act 2
Solution



The work-energy theorem says that for any object WNET = K
In this example the only force that does work is friction (since
both N and mg are perpendicular to the blocks motion).
The net work done to stop the box is - fD = -mmgD.
 This work “removes” the kinetic energy that the box had:
 WNET = K2 - K1 = 0 - K1
m
D
Physics 111: Lecture 9, Pg 23
Lecture 9, Act 2
Solution


The net work done to stop a box is - fD = -mmgD.
This work “removes” the kinetic energy that the box had:
WNET = K2 - K1 = 0 - K1
This is the same for both boxes (same starting kinetic energy).
mm2gD2  mm1gD1
Since m1 > m2 we can see that
m2D2  m1D1
D2 > D1
m1
D1
m2
D2
Physics 111: Lecture 9, Pg 24
A simple application:
Work done by gravity on a falling object


What is the speed of an object after falling a distance H,
assuming it starts at rest?
Wg = F r = mg r cos(0) = mgH
v0 = 0
r
Wg = mgH
mg
j
H
Work/Kinetic Energy Theorem:
Wg = mgH = 1/2mv2
v  2 gH
v
Physics 111: Lecture 9, Pg 25
What about multiple forces?
Suppose FNET = F1 + F2 and the
displacement is r.
The work done by each force is:
W1 = F1 r
W2 = F2  r
F1
FNET
WTOT = W1 + W2
= F1 r + F2 r
= (F1 + F2 ) r
WTOT = FTOT r
r
F2
It’s the total force that matters!!
Physics 111: Lecture 9, Pg 26
Comments:

Time interval not relevant
Run up the stairs quickly or slowly...same W
Since W = F r

No work is done if:
F = 0
or
r = 0
or
 = 90o
Physics 111: Lecture 9, Pg 27
Comments...
W = F  r

No work done if  = 90o.
T
No work done by T.
v
v
No work done by N.
N
Physics 111: Lecture 9, Pg 28
Lecture 9, Act 3
Work & Energy

An inclined plane is accelerating with constant acceleration a.
A box resting on the plane is held in place by static friction.
How many forces are doing work on the block?
a
(a) 1
(b) 2
(c) 3
Physics 111: Lecture 9, Pg 29
Lecture 9, Act 3
Solution

First, draw all the forces in the system:
FS
a
mg
N
Physics 111: Lecture 9, Pg 30
Lecture 9, Act 3
Solution

Recall that W = F Δr so only forces that have a
component along the direction of the displacement are
doing work.
FS
a
mg

N
The answer is (b) 2.
Physics 111: Lecture 9, Pg 31
Recap of today’s lecture






Work & Energy
Discussion
Definition
Dot Product
Work of a constant force
Work/kinetic energy theorem
Properties (units, time independence, etc.)
Work of a multiple forces
Comments
Physics 111: Lecture 9, Pg 32
Physics 111: Lecture 10
Today’s Agenda





Review of Work
Work done by gravity near the Earth’s surface
Examples:
pendulum, inclined plane, free fall
Work done by variable force
Spring
Problem involving spring & friction
Physics 111: Lecture 9, Pg 33
Review: Constant Force
Work, W, of a constant force F
acting through a displacement r
is:
W = F r = F r cos() = Fr r
F

Fr
r
Physics 111: Lecture 9, Pg 34
Review: Sum of Constant Forces
Suppose FNET = F1 + F2 and the
displacement is S.
The work done by each force
is:
W1 = F1 r
W2 = F2 r
WNET = W1 + W2
= F1 r + F2 r
= (F1 + F2 ) r
WNET = FNET  r
F1
FTOT
r
F2
Physics 111: Lecture 9, Pg 35
Review: Constant Force...
W = F r

No work done if  = 90o.
T
No work done by T.
v
v
N
No work done by N.
Physics 111: Lecture 9, Pg 36
Work/Kinetic Energy Theorem:
{Net Work done on object}
=
{change in kinetic energy of object}
WF = K = 1/2mv22 - 1/2mv12
v1
v2
F
m
WF = Fx
x
Physics 111: Lecture 9, Pg 37
Work done by gravity:

Wg = F r = mg r cos 
= -mg y
m
mg
Wg = -mg y
r 
j
y
Depends only on y !
m
Physics 111: Lecture 9, Pg 38
Work done by gravity...

W NET = W1 + W2 + . . .+ Wn
= F r 1+ F r2 + . . . + F rn
= F (r1 + r 2+ . . .+ rn)
= F r
= F y
m
r1
y
r3
Wg = -mg y
Depends only on y,
not on path taken!
r
mg
r2
j
rn
Physics 111: Lecture 9, Pg 39
Lecture 10, Act 1
Falling Objects

Falling
objects
Three objects of mass m begin at height h with velocity 0. One
falls straight down, one slides down a frictionless inclined
plane, and one swings on the end of a pendulum. What is the
relationship between their velocities when they have fallen to
height 0?
v=0
v=0
v=0
H
vf
Free Fall
(a) Vf > Vi > Vp
vi
Frictionless incline
(b) Vf > Vp > Vi
vp
Pendulum
(c) Vf = Vp = Vi
Physics 111: Lecture 9, Pg 40
Lecture 10, Act 1
Solution
v=0
v=0
v=0
H
vf
Free Fall
vi
vp
Frictionless incline
Pendulum
Only gravity will do work: Wg = mgH = 1/2 mv22 - 1/2 mv12 = 1/2 mv22
v f  v i  v p  2 gH
does not depend on path !!
Physics 111: Lecture 9, Pg 41
Lifting a book with your hand:
What is the total work done on the book??

First calculate the work done by gravity:
Wg = mg r = -mg r

Now find the work done by
the hand:
WHAND = FHAND r = FHAND r
r
FHAND
v = const
a=0
mg
Physics 111: Lecture 9, Pg 42
Example: Lifting a book...
Wg
= -mg r
WHAND = FHAND r
r
WNET
Textbook
FHAND
v = const
a=0
= WHAND + Wg
= FHAND r - mg r
= (FHAND - mg) r
mg
= 0 since ΔK = 0 (v = const)

So WTOT = 0!!
Physics 111: Lecture 9, Pg 43
Example: Lifting a book...

Work/Kinetic Energy Theorem says: W = K
{Net Work done on object} = {change in kinetic energy of object}
In this case, v is constant so K = 0
and so W must be 0, as we found.
r
FHAND
v = const
a=0
mg
Physics 111: Lecture 9, Pg 44
Work done by Variable Force: (1D)

When the force was constant, we
wrote W = F x
area under F vs. x plot:
F
Wg
x
x

For variable force, we find the area
by integrating:
dW = F(x) dx.
F(x)
x2
W   F ( x )dx
x1
x1
dx
x2
Physics 111: Lecture 9, Pg 45
Work/Kinetic Energy Theorem for a
Variable Force
x2
W   F dx
x1
x2
m
x1
v2
dv dx
dt
mv
v1
F  ma  m dv
dt
dv dx dv
dv
=
= v dx (chain rule)
dt
dt dx
dv
dx
dx
v2
 m  v dv
v1
1
1
1
 m (v22 v12 )  m v22  mv12  ΔKE
2
2
2
Physics 111: Lecture 9, Pg 46
1-D Variable Force Example: Spring

For a spring we know that Fx = -kx.
F(x)
x1
x2
x
relaxed position
-kx
F = - k x1
F = - k x2
Physics 111: Lecture 9, Pg 47
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.
F(x)
relaxed position
x1
x2
x
Ws
-kx
Physics 111: Lecture 9, Pg 48
Spring...

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
F(x)
Ws   F ( x )dx
x1
x1
x2
x2
x
Ws
-kx
  ( kx )dx
x1
1
  kx 2
2
Ws  
x2
x1
1
k x22  x12 
2
Physics 111: Lecture 9, Pg 49
Lecture 10, Act 2
Work & Energy

A box sliding on a horizontal frictionless surface runs into a
fixed spring, compressing it a distance x1 from its relaxed
position while momentarily coming to rest.
If the initial speed of the box were doubled and its mass
were halved, how far x2 would the spring compress ?
(a)
x2  x1
(b) x2  2 x1
(c)
x2  2 x1
x
Physics 111: Lecture 9, Pg 50
Lecture 10, Act 2
Solution

Again, use the fact that WNET = K.
WNET = WSPRING = -1/2 kx2
K = -1/2 mv2
In this case,
and
so
kx2 =
mv2
v1
In the case of x1
x1  v1
m1
k
x1
m1
m1
Physics 111: Lecture 9, Pg 51
x v
m
k
Lecture 10, Act 2
Solution
So if v2 = 2v1 and m2 = m1/2
x 2  2v 1
m1 2
k
 v1
2m1
k
x2  2 x1
v2
x2
m2
m2
Physics 111: Lecture 9, Pg 52
Problem: Spring pulls on mass.

A spring (constant k) is stretched a distance d, and a mass m
is hooked to its end. The mass is released (from rest). What
is the speed of the mass when it returns to the relaxed
position if it slides without friction?
m
relaxed position
m stretched position (at rest)
d
m
after release
v
m
back at relaxed position
vr
Physics 111: Lecture 9, Pg 53
Problem: Spring pulls on mass.

First find the net work done on the mass during the motion
from x = d to x = 0 (only due to the spring):
Ws  
1
1
1
k x22  x12    k 0 2  d 2   kd 2
2
2
2
m stretched position (at rest)
d
m
vr
relaxed position
i
Physics 111: Lecture 9, Pg 54
Problem: Spring pulls on mass.

Now find the change in kinetic energy of the mass:
1
1
1
ΔK  mv 22  mv12  mv r2
2
2
2
m stretched position (at rest)
d
m
vr
relaxed position
i
Physics 111: Lecture 9, Pg 55
Problem: Spring pulls on mass.

Now use work kinetic-energy theorem: Wnet = WS = K.
1
kd 2 
2
1
mv r 2
2
vr  d
k
m
m stretched position (at rest)
d
m
vr
relaxed position
i
Physics 111: Lecture 9, Pg 56
Problem: Spring pulls on mass.


Now suppose there is a coefficient of friction m between the
block and the floor
The total work done on the block is now the sum of the work
done by the spring WS (same as before) and the work done by
friction Wf.
.
Wf = f Δr = - mmg d
r
m stretched position (at rest)
d
m
vr
f = mmg
relaxed position
i
Physics 111: Lecture 9, Pg 57
Problem: Spring pulls on mass.

Again use Wnet = WS + Wf = K
1
2
W

kd
S
Wf = -mmg d
2
1 2
1
2
kd  mmgd  mv r
2
2
1
2
K  mv r
2
vr 
k 2
d  2 μgd
m
r
m stretched position (at rest)
d
m
vr
f = mmg
relaxed position
i
Physics 111: Lecture 9, Pg 58
Recap of today’s lecture





Review
Work done by gravity near the Earth’s surface
Examples:
pendulum, inclined plane, free fall
Work done by variable force
Spring
Problem involving spring & friction
Physics 111: Lecture 9, Pg 59