Download Physics 5

Physics
Chapter 5:
Work & Energy
Work and Energy
• Work
– A Force that Causes the
Displacement of an Object Does
Work on the Object
Work and Energy
• Work
–Has Direction (+ or -)
• May Increase Energy of the
System (+)
• May Decrease Energy of the
System (-)
W  Fd
Work and Energy
• Work
–Units
• Force – Newton (kg*m/s2)
• Displacement – Meter
• Work – Joule (kg*m2/s2) – (Nm)
W  Fd
Work and Energy
• Work
– A Horizontal Component of the
Force (Relative to Motion) Must Do
the Work
W  Fd
Work and Energy
• Work
– Work Done at an Angle
• Forces Only Move in One Direction, but Involve 2
Axis
• These Axis Vectors Can Be Separated and
Calculated Individually
Y+
F
X+
Work and Energy
• Work
– Work Done at an Angle
Y+
F
Fy
q
X+
Fx
Work and Energy
• Work
– Work Done at an Angle
Fx  F cos q
Y+
F
q
X+
Fx
Work and Energy
• Work
– Work Done at an Angle
Fy  F sin q
Y+
q
F
Fy
q
X+
Work and Energy
• Work
– Work Done at an Angle
• May be Negative
Fx  F cos q
Fy   F sin q
Y+
Fx
X+
q
Fy
F
Work and Energy
W  Fd
• Work
Wx  Fx d  F (cos q )d
– Work Done at an Angle
• To Calculate Work
Wy  Fy d   F (sin q )d
Y+
Fx
X+
q
Fy
F
Wnet  Wx  W y
Work and Energy
• Work
• Problem “Work Truck”
– m = 2800kg
– F = 29N
– d = 500m
• What is W?
Y+
d
X+
Work and Energy
• Work
• Solution
– m = 2800kg
– F = 29N
– d = 500m
W  Fd  (29 N )(500m)  1.4 x10 J
4
Y+
d
X+
Work and Energy
• Work
• Problem “Building Moving”
– m = 6600kg
– d = 2.5m
• What is the Amount of Work?
Work and Energy
• Work
• Solution “Building Moving”
– m = 6600kg
– d = 2.5m
F  ma  (6600kg)(9.8m / s )  6.5x10 N
2
4
Work and Energy
• Work
• Solution “Building Moving”
– m = 6600kg
– d = 2.5m
– F = 6.5x104N
W  Fd  (6.5x10 N )(2.5m)  1.6 x10 J
4
5
Work and Energy
• Work
• Problem
– The third floor of a house is 8m above street level.
How much work is needed to move a 150kg
refrigerator to the third floor?
Work and Energy
• Work
• Solution
– d = 8m
– m = 150kg
F  ma  (150kg)(9.8m / s )  1.47 x10 N
2
3
W  Fd  (1.47 x10 N )(8m)  1.1x10 J
3
4
Work and Energy
• Work
• Problem
– John pushes a box across the floor of a stage with
a horizontal force. The roughness of the floor
changes, and John must exert a force of 20N for
5m, then 35N for 12m, then 10N for 8m. What is
the work that John has done pushing the box?
Work and Energy
• Work
• Solution
– d1 = 5m
– F1 = 20N
– d2 = 12m
– F2 = 35N
– d3 = 8m
– F3 = 10N
W  (Fd )
Work and Energy
• Work
• Solution
W  F1d1  F2 d 2  F3d3
W  (20 N )(5m)  (35 N )(12m)  (10 N )(8m)  600 J
Work and Energy
• Work
• Problem
– A pump delivers 0.55m3 of oil into barrels on a
platform 25.0m above the pump intake pipe. The
density of the oil is 0.82g/cm3. What is the work
done by the pump?
Work and Energy
• Work
• Solution
– d = 25.0m
– volume = 0.55m3
– density = 0.82g/cm3
– mass = ?
5
0.82 g " x" g 8.2 x10 g
2
3
 3 
 8.2 x10 kg / m
3
3
cm
m
m
Work and Energy
• Work
• Solution
– d = 25.0m
– volume = 0.55m3
– density = 0.82g/cm3
– mass = ?
(8.2 x102 kg / m3 )(0.55m3 )  4.5x102 kg
Work and Energy
• Work
• Solution
– d = 25.0m
– volume = 0.55m3
– density = 0.82g/cm3
– mass = 4.5x102kg
–F=?
Fg  mg  (4.5x10 kg)(9.8m / s )  4.4 x10 N
2
2
3
Work and Energy
• Work
• Solution
– d = 25.0m
– volume = 0.55m3
– density = 0.82g/cm3
– mass = 4.5x102kg
– F = 4.4x103N
W  Fd  (4.4 x103 N )(25.0m)  1.1x105 J
Work and Energy
• Problem
– Ryan pulls his race car with a rope from his
personal car. He accelerates from rest to a speed
of 25km/hr in 300m. The angle of the rope is
15o. The race car’s mass is 682kg. What is the
force on the rope?
Y+
Fy
F
q
X+
Fx
Work and Energy
• Solution
–
–
–
–
–
–
25km/hr
d = 300m
q = 150
m = 682kg
Fx = ?
F=?
Work and Energy
• Solution
–
–
–
–
–
–
–
vo = 0m/s
vf = 6.9m/s
d = 300m
q = 150
m = 682kg
Fx = ?
F=?
25km / hr  2.5 x10 m / 3.6 x10 sec  6.9m / s
4
3
Work and Energy
• Solution
–
–
–
–
–
–
–
vi = 0m/s
vf = 6.9m/s
d = 300m
q = 150
m = 682kg
Fx = ?
F=?
1 / 2mv f  1 / 2mvi  Fd
2
2
1 / 2mv f  1 / 2mvi
2
Fx 
2
d
1 / 2(682kg)(6.9m / s) 2  1 / 2(682kg)(0m / s) 2
Fx 
 54 N
300m
Work and Energy
• Solution
–
–
–
–
–
–
–
vi = 0m/s
vf = 6.9m/s
d = 300m
q = 150
m = 682kg
Fx = 54N
F=?
Fx
54 N
F


55
.
9
N
0
cos q cos 15
Work and Energy
• Problem
– Ryan pulls his race car with a rope from his
personal car. He accelerates from rest to a speed
of 25km/hr in 300m. The angle of the rope is
150. The race car’s mass is 682kg. What is the
work done?
Y+
Fy
F
q
X+
Fx
Work and Energy
• Solution
–
–
–
–
–
–
–
vi = 0m/s
vf = 6.9m/s
d = 300m
q = 150
m = 682kg
Fx = 54N
F = 55.9N
Wx  Fx d  ( F ) cosq (d )
Wx  Fx d  (54 N ) cos15 (300m)  1.6 x10 J
0
4
Work and Energy
• Solution
–
–
–
–
–
–
vi = 0m/s
vf = 6.9m/s
d = 300m
q = 150
m = 682kg
F = 54N
Wy  Fy d  ( F ) sin q (d )
Wy  Fy d  (14.5N ) sin 15 (300m)  1.1x10 J
0
3
Work and Energy
• Solution
–
–
–
–
–
–
Wnet  Wx  W y
vi = 0m/s
vf = 6.9m/s
d = 300m
q = 150
m = 682kg
F = 54.6N
Wnet  1.6 x10 J  1.1x10 J  1.7 x10 J
4
3
4
Work and Energy
• Energy
– The Ability to Produce Change
Within a System
• Thermal Energy
• Chemical Energy
• Energy of Motion
Work and Energy
• Energy
– Energy of Motion
• The Energy of an Object Resulting from
Motion
• Kinetic Energy
Work and Energy
• Energy
– Kinetic Energy
• Kinetics
• Newton’s Second Law of Motion
F  ma
Work and Energy
• Energy
– Kinetic Energy
• Kinetics
• Motion Equation
–
–
–
–
Vi is initial velocity
Vf is the final velocity
a is the instantaneous acceleration
d is the displacement
v f  vi  2ad
2
2
Work and Energy
• Energy
– Kinetic Energy
v f  vi  2ad
2
• Kinetics
2
F
a
m
F  ma
vf
2
Fd
 vi  2
m
2
Work and Energy
• Energy
– Kinetic Energy (K)
• Kinetics
vf
2
Fd
 vi  2
m
2
1 / 2mv f  1 / 2mvi  Fd
2
2
Work and Energy
• Energy
– Kinetic Energy (K)
• Kinetics
Energy of the System = Environment
1 / 2mv f  1 / 2mvi  Fd
2
2
KE  1 / 2mv
2
Work and Energy
• Energy
– Kinetic Energy (K)
• Kinetics
System (Change in Energy) = Environment (Object)
1 / 2mv f  1 / 2mvi  Fd
2
2
A Change in the Environment Due to Energy = Work
W  Fd
Work and Energy
• Energy
– Kinetic Energy (K)
• Kinetics
A Change in the Environment Due to Energy = Work
W  Fd
KE  W
Work and Energy
• Energy
– Kinetic Energy (K)
• Kinetics
– Work-Energy Theorem
KE  W
Work and Energy
• Energy
– Kinetic Energy
• Units
– Joule (J)
– 1J=1kg*m2/s2
KE  W
Work and Energy
• Work
• Problem “Work Truck”
– m = 2800kg
– F = 29N
– d = 500m
• What is KE?
Y+
d
X+
Work and Energy
• Work
• Solution
– m = 2800kg
– F = 29N
– d = 500m
KE  W  1.4 x10 J
4
Y+
d
X+
Work and Energy
• Energy
– Potential Energy (Gravitational)
• Any Object Held by a Normal Force
Against the Force of Gravity Contains
Potential Energy
Work and Energy
• Energy
– Potential Energy (Gravitational)
• The product of…
– the Mass of the Object
– the Acceleration of Gravity
– the Height of the Object
PEg  mgh
Work and Energy
• Energy
– Potential Energy (Elastic)
1 2
PEe  kx
2
Work and Energy
• Homework
– Page 193 - 194
• Problems
– 7 (53J, ?)
– 9 (47.5J)
– 11 (?, ?, ?)
– 23 (a, 5400J, 0J, 5400J b, 0J, -5400J, 5400J c,
2700J, -2700J, 5400J)
– 25 (a, 0.4J b, 0.225J, c, 0J)
Work and Energy
• Energy
– Conservation of Energy
• An Object’s Potential Energy is Equal to its Kinetic
Energy When the Normal Force is Released
PE  KE
Work and Energy
• Energy
– Conservation of Energy
PE  mgh
1 2
KE  mv
2
PE  KE
1 2
mgh  mv
2
Work and Energy
• Energy
– Conservation of Energy
1 2
mgh  mv
2
1 2
gh  v
2
Work and Energy
• Energy
– Conservation of Energy
1 2
W  Fd  KE  mv  PE  mgh
2
Work and Energy
• Power
– Work Can be Done Quickly, or…
– The Same Work Can be Done Over a
Long Period of Time
– Power is the Rate of Doing Work (the rate
of energy transferred)
Work and Energy
• Power (P)
W = Work (Joules)
t =Time (sec)
“The Less Time,
the Greater the
Power”
W
P
t
Work and Energy
• Power (P)
Units are Watts
1 watt = 1 J/s
W
P
t
Work and Energy
• Power (P)
Problem: “Lift that Barge”
Ryan Hauls His Piano Up Two Flights of
Stairs (22 m) in 5 minutes. The Piano’s
mass is 90.7 kg. What Power is Used?
W
P
t
Work and Energy
• Power (P)
Solution: “Lift that Barge”
d = 22m
m = 90.7kg
t = 3x102s
W
P
t
Work and Energy
• Power (P)
Solution: “Lift that Barge”
d = 22m
m = 90.7kg
t = 3x102s
F  ma  (90.7kg)(9.8m / s)  8.9 x10 N
2
Work and Energy
• Power (P)
Solution: “Lift that Barge”
d = 22m
m = 90.7kg
t = 3x102s
F = 8.9x102N
W
P
t
W  Fd
Work and Energy
• Power (P)
Solution: “Lift that Barge”
d = 22m
m = 90.7kg
t = 3x102s
F = 8.9x102N
Fd
P
t
2
(8.9 x10 N )( 22m)
P

65
w
2
3x10 s
Work and Energy
• Power (P)
Problem: “Lift that Barge II”
What If Ryan Hustled the Piano Up the
Stairs in 20 seconds?
Work and Energy
• Power (P)
Solution: “Lift that Barge”
d = 22m
m = 90.7kg
t = 20s
F = 8.9x102m
W
P
t
Work and Energy
• Power (P)
Solution: “Lift that Barge”
d = 22m
m = 90.7kg
t = 20s
F = 8.9x102m
2
W Fd (8.9 x10 N )( 22m)
P 

 979w
t
t
20s
Work and Energy
• Power (P)
WOW!
To Move the Piano in 5 Minutes – 65w
To Do the Same Work in 20 seconds – 979w
Yes, 20 seconds is 1/15 of the original 5 minute time and
65.3 x 15 = 979. Hey! A Direct (linear) Relationship!
Work and Energy
• Problem
– A force of 300.0 N is used to push a 145-kg
mass 30.0 m horizontally in 3.00 s. What
is the power developed?
Work and Energy
• Solution
– F = 300N
– d = 30.0m
– m = 145kg
– t = 3.0s
W  Fd  (300 N )(30.0m)  9 x103 J
3
W 9 x10 J
P 
 3kw
t
3.0s
Work and Energy
• Problem
– April pushes a wheelbarrow by exerting a
145N force horizontally. April moves it
60.0m at a constant speed for 25.0s. What
power does April develop?
Work and Energy
• Solution
– F = 145N
– d = 60.0m
– t = 25.0s
W Fd (145 N )(60.0m)
P


 348w
t
t
25.0s
Work and Energy
• Problem
– If April moves the wheelbarrow twice as
fast, how much power is developed?
Work and Energy
• Solution
– F = 145N
– d = 60.0m
– t = 12.5s
W Fd (145 N )(60.0m)
P


 696w
t
t
12.5s
Work and Energy
• Problem
– Jon pulls a 305N sled along a snowy path
using a rope that makes a 45.0° angle with
the ground. Jon pulls with a force of
42.3N. The sled moves 16m in 3.0s. What
power does Jon produce?
Work and Energy
• Solution
– Fg = 305N
– F = 42.3N
– d = 16.0m
– t = 3.0s
– q = 450
W Fd cos q (42.3N )(16.0m)(cos 450 )
P 

 160w
t
t
3.0s
Work and Energy
• Problem
– A lawn roller is pushed across a lawn by a
force of 115N along the direction of the
handle, which is 22.5° above the
horizontal. If you use 64.6w of power for
90.0s, what distance is the roller pushed?
Work and Energy
• Solution
– F = 115N
– P = 64.6w
– t = 90.0s
– q = 22.50
Pt
(64.6w)(90.0s)
d

 54.7m
0
F cos q (115 N )(cos 22.5 )
Work and Energy
• Problem
– You slide a crate up a ramp at an angle of
30.0° by exerting a 225N force parallel to
the ramp. The crate moves at constant
speed. The coefficient of friction is 0.28.
How much work have you done on the
crate when it is raised a vertical distance
of 1.15m?
Work and Energy
• Solution
– F = 225N
– mk = 0.28
– h = 1.15m
– q = 30.00
h
1.15m
d

 2.3m
0
sin q (sin 30.0 )
W  Fd  (225 N )( 2.3m)  518 J
Work and Energy
• Problem
– An engine moves a boat through the water
at a constant speed of 15 m/s. The engine
must exert a force of 6.0x103N to balance
the force that water exerts against the
hull. What power does the engine
develop?
Work and Energy
• Solution
– F = 6.0x103N
– v = 15m/s
W Fd
d
P

 F  Fv
t
t
t
P  (6.0 x10 N )(15m / s)  90kw
3
Work and Energy
• Problem
– A 188W motor will lift a load at the rate
(speed) of 6.50cm/s. How great a load can
the motor lift at this rate?
Work and Energy
• Solution
– P = 188W
– v = 6.5cm/s = 0.065m/s
W Fd
d
P

 F  Fv
t
t
t
P
188W
3
F 
 2.89 x10 N
v 0.065m / s
Work and Energy
• Homework:
– Pages 195 – 197
• Problems
– 33 (12.0m/s)
– 35 (17.2s)
– 49 (a, 2.25x104N b, 1.33x10-4s)
– 57 (a, 4.4m/s b, 1.5x105N)