Download Lecture 16

Seating for PHYS 1020 Midterm
Thursday, October 22
7 - 9 pm
Seating is by last name
Room
From
To
111 Armes
A
BJ
200 Armes
BL
GA
201 Armes
GH
KH
204 Armes
KI
OB
205 Armes
OK
SA
208 Armes
SC
Z
20 multiple choice questions, ch 1-5. Formula sheet provided.
Monday, October 19, 2009
40
WileyPLUS Assignment 2
Due today at 11:00 pm !
Chapters 4 & 5
This Wednesday:
Review of ch 1 - 5 for the midterm
Email questions from old exams (back of lab manual),
from the book, or WileyPLUS !
Monday, October 19, 2009
41
Conservation of Mechanical Energy
Mechanical energy = KE + PE = mv2/2 + mgy
In the absence of applied forces and friction:
(that is, non-conservative forces)
(change in KE) + (change in PE) = 0
so mechanical energy is conserved.
KE + PE = constant
When an applied force does work, the work-energy theorem becomes:
W = ΔKE + ΔPE
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42
Power
Power is the rate of doing work, or the rate at which energy
is generated or delivered.
Unit: 1 watt (W) = 1 J/s
Power, P =
W Fs
s
=
= F × = Fv
t
t
t
(speed = distance/time)
v
P = Fv
m
F
a
s
Kilowatt-hour (kWh): the energy generated or work done
when 1 kW of power is supplied for 1 hour.
1 kWh = (1000 J/s)!(3600 s) = 3,600,000 J = 3.6 MJ
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43
6.82/74: In 2 minutes, a ski lift raises 4 skiers at constant speed
to a height of 140 m. The average mass of each skier is 65 kg.
What is the average power provided by the tension in the cable
pulling the lift?
Consider the power needed to
lift 1 skier of mass 65 kg.
T
h = 140 m
KE is constant, PE increases
m = 65 kg
Work done by the applied force, T, is:
(work-energy theorem)
Wnc = ∆KE + ∆P E = 0 + mgh
The power needed = work/time
P =
Wnc
mgh
65 × g × 140
=
=
= 740 W
t
t
120
per skier
Total = 2960 W
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44
6.-/60: A motorcycle (mass of cycle + rider = 250 kg) is travelling
at a steady speed of 20 m/s. The force of air resistance on cycle
+ rider is 200 N. Find the power necessary to maintain this speed
if a) the road is level and b) slopes upward at 370.
a)
F�r
�
F
�v
Work-energy theorem: Wnc = ΔKE + ΔPE,
and ΔKE = ΔPE = 0
The force supplied by the engine F = Fr = 200 N
Power needed, P = Fv = 200 ! 20 = 4000 W
b)
�r
F
�
F
! = 370
Monday, October 19, 2009
�v
(5.4 hp)
The motorcycle gains
potential energy, so an extra
amount of energy must be
supplied by the engine.
45
�
F
�r
F
b)
!=
�v
h = v sin! in 1 s
370
Work-energy theorem: Wnc = ΔKE + ΔPE,
and ΔKE = 0
In 1 s, cycle goes up an amount h = v sinθ
(travels distance v in 1 s)
So, extra work done by engine in 1 s is given by ΔPE = mgv sinθ
So, P = 4000 + mgv sinθ
= 4000 + 250 ! g ! 20 sin370
= 33,500 W
(45 hp)
Monday, October 19, 2009
46
6.70/62: A 1900 kg car travels up and down a hill at a constant
speed of 27 m/s. The force of air resistance and friction is of
the same magnitude in both directions.
Going up the hill, the car’s engine needs to produce 47 hp (35
kW) greater power to maintain speed than it does going down the
same hill.
What is the angle of inclination of the hill?
v = 27 m/s
Fr
!
P! – P = 35 kW (47 hp)
Power P!
v = 27 m/s
Going up
Fr
Power P
!
Going down
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47
Other Forms of Energy
There are many forms of energy:
• Electrical
• Elastic (eg energy stored in a spring)
• Chemical
• Thermal
• Nuclear
Energy is conserved overall:
Energy may be converted from one form to another, but the
total amount of energy is conserved.
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48
Work done by a variable force
Example: compound bow
– a number of pulleys and
strings
• maximize the energy
stored in the bow for
finite effort
F
• reduced force with
bow fully drawn.
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49
Force to draw the bow
Reduced
effort needed
when bow fully
drawn
Displacement, s
How much work is needed to draw the bow?
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50
Work done is force ! distance...
Split the curve into segments
W � (F cos !)1"s1 + (F cos !)2"s2 + . . .
= sum of force " distance
Becomes exactly the area under the curve when the slices become
vanishingly narrow → integral calculus
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51
Work done in pulling
back the bowstring
Work done in drawing the
bow = area under the curve
Count the squares, multiply by
area of each.
Number of squares under the curve
" 242.
Area of each square is:
(9 N) ! (0.0278 m)
= 0.25 N.m = 0.25 J.
So, work done is W = 242 ! 0.25 = 60.5 J
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52
6.72/66
Work done = area under triangular curve
1
= × (base) × (height)
2
W = 0.5 ! (1.6 m) ! (62 N) = 49.6 J
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53
6.-/67
A force is applied
to a 6 kg mass
initially at rest.
a) How much
work is done by
the force?
b) What is the
speed of the mass
at s = 20 m?
a) Work done = area under the force-displacement curve
1
W = × (10 m) × (10 N) + (20 − 10 m)(10 N) = 150 J
2
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54
b) What is the speed of the mass at s = 20 m?
Wnc = !KE + !PE = mv2/2 + 0 = 150 J
v=
Monday, October 19, 2009
�
2Wnc/m =
�
2 × 150/6 = 7.07 m/s
55
Summary
In absence of non-conservative forces:
Conservation of mechanical energy: E = KE + PE = constant
When non-conservative forces are present:
Work-energy theorem: Wnc = ΔKE + ΔPE
Power = rate of doing work (1 W = 1J/s)
P = Fv
Work done by a variable force = area under the force versus
displacement curve
Monday, October 19, 2009
56
5.38/-: Two newly discovered planets follow circular orbits
around a star in a distant part of the galaxy. The orbital speeds
of the planets are determined to be 43.3 km/s and 58.6 km/s.
The slower planet’s orbital period is 7.6 years.
a) What is the mass of the star?
b) What is the orbital period of the faster planet?
Monday, October 19, 2009
57
4.93/109: The coefficient of kinetic friction between the block and
the table is 0.1.
a) What is the acceleration of the three blocks?
b) Find the tension in the two strings.
�a
FN
T1
T2
Fk
T1
m2g
T2
�a
�a
m1g
m3g
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58
3.50/54: A plane is to fly due west. The plane flies at 245 m/s
relative to the air. A 38 m/s wind is blowing from south to north. In
what direction should the plane head in order to travel due west?
The resultant velocity vector should point to the west. It is the sum
of the velocity of the plane relative to the ground, vp, and of the
wind relative to the ground, vw.
Resultant velocity:
245 m/s
vp
v = vw + vp
38 m/s
vw
θ
v = resultant
sin θ =
vw
38
=
→ θ = 8.9◦ S of W
vp
245
Monday, October 19, 2009
59