Transcript Practice - Mr. Lee at ASGL

Honors Physics Chapter 10
Work, Energy, and
Simple Machine
1
Next Generation Science Standards
HS-PS3-1.
Create a computational model to calculate the
change in the energy of one component in a
system when the change in energy of the other
component(s) and energy flows in and out of the
system are known.
HS-PS3-2.
Develop and use models to illustrate that energy at
the macroscopic scale can be accounted for as
either motions of particles or energy stored in
fields.
2
Honors Physics



3
Turn in Chapter 9 homework, worksheet, and
Lab Report
Lecture
Q&A
Work, W
Work: the product of the force exerted on an
object and the distance (displacement) the
object moves in the direction of the force.
W  Fd


Valid only when F and d
are in the same direction.
F: force
d: displacement, more exactly, d
Unit:
[W] = [F] · [d] = N · m = Joule = J
4
Example: Pg261pp2
Together, two students exert a force of 825 N in pushing a car a
distance of 35 m.
a) How much work is done?
b) If the force was doubled, how much work would they do pushing
the car the same distance?
F  825 N , d  35m
a )W  ?
4
W  Fd  825N  35m  28875J  2.9 10 J
b) F  2  825 N  1650 N , d  35m, W  ?
W  Fd  1650 N  35m  5.8 104 J
5
Practice: Pg261pp3
A rock climber wears a 7.5-kg backpack while scaling a cliff. After 30.0
min, the climber is 8.2 m above the starting point.
a) How much work does the climber do on the backpack?
b) If the climber weighs 645 N, how much work does she do lifting
herself and the backpack?
mb  7.5kg , t  30.0 min, d  8.2m
a)Wb  ?
F  mb g  7.5kg  9.8
m
 73.5 N
s2
W  Fd  73.5 N  8.2m  602.7 J  603J
b) Fg  645 N , Wg b  ?
F  73.5N  645N  718.5N
W  Fd  718.5 N  8.2m  5891.7 J  5890 J
6
Practice
What work is done by a forklift raising a 583-kg
box 1.2 m?
m  583kg , d  1.2m,W  ?
m
F  mg  583kg  9.8 2  5713N
s
W  Fd  5713N 1.2m  6856 J  6900 J  6.9kJ
7
Practice:
If you push twice as hard against a stationary
brick wall, the amount of work you do
a) doubles.
b) is cut in half.
c) remains constant but non-zero.
d) remains constant at zero.
8
What if
F and d are not in the same direction?
W  Fd cos 
F: force
 d: displacement (more accurately, d)
 : angle between F and d

9
Positive and Negative Work
W  Fd cos



<
 W ____
> 0
 = 90o  W ____
= 0
 > 90o  W ____
< 0
90o

F=0
d=0
 = 90o
10
W=0
Work is a scalar—it has no
direction.
Negative sign does not indicate
direction.
A positive work is always larger
than a negative work.
Example: Pg262pp5
Two people lift a heavy box a distance of 15 m. They use
ropes, each of which makes an angle of 15o with the
vertical. Each person exerts a force of 225 N. How much
work do they do?
d  15m,   15o , F  2  225N  450 N , W  ?
d
W  Fd cos   450 N  15m  cos15o  6520 J
11

F
Practice: Pg262pp7
A rope is used to pull a metal box 15.0 m across the floor.
The rope is held at an angle of 46.0o with the floor and a
force of 628 N is applied to the rope. How much work
does the force on the rope do?
F
46.0o
d  15.0m,  46.0o , F  628N ,W  ?
W  Fd cos   628 N 15.0m  cos 46.0o  6540 J
12
Practice:
If you walk 5.0 m horizontally forward at a
constant velocity carrying a 10-N book, the
amount of work you do is
a) more than 50 J.
b) equal to 50 J.
c) less than 50 J, but more than 0 J.
d) zero.
13
Work Done by Weight (Gravity)
When an object goes up or down, the work
done by weight (gravity) is given by
W  mg h
W  Fd cos   mgd cos 
still works.
where
 m: mass of object
 g = 9.8 m/s2
 Δh: change in height of object


14

< 0
Δh > 0 when going up  W ____
> 0
Δh < 0 when going down  W ____
Δh = 0 when going horizontally  W ____
= 0
Practice: Pg262-8
A bicycle rider pushes a bicycle that has a mass of 13 kg up a steep hill.
The incline is 25o and the road is 275 m long, as shown in Figure 104. The rider pushes the bike parallel to the road with a force of 25 N.
a.
How much work does the rider do on the bike?
b.
How much work is done by the force of gravity on the bike?
m  13kg , d  275m, F  25N ,  25o
a) Fd  0 ,W  ?
o

gd

W  Fd cosFd  25N  275m  cos 0o  6875J
b)Wg  ?
h  d sin   275m  sin 25o  116m
m
Wg  mg h  13kg  9.8 2  116m  1.48  104 J
s
Fg  mg
h
Fg
 gd  90o    90o  25o  115o
Wg  Fg d cos  gd  mgd cos  gd  13kg  9.8
15
m
 275m  cos115o  1.48  104 J
2
s
Practice:
You lift a 10-N physics book up in the air a
distance of 1.0 m, at a constant velocity of
0.50 m/s. What is the work done by the
weight of the book?
a) 10J
b) -10 J
c) 5.0 J
d) -5.0 J
16
Power, P
Power: (time) rate of doing work, or (time) rate of
energy transfer
Average power:


or
P
E
t
W: Work done
t: time, duration, for the work done (more exactly, Δt)
Unit:
17
W
P
t
W
J


 P 
t

s
 Watt  W
1 horse power = 1 hp = 746 W
Instantaneous Power
P  Fv cos
•
•
•
18
F: force
v: velocity
: angle between F and v
Example: Pg264pp9
A box that weighs 575 N is lifted a distance of 20.0 m
straight up by a rope. The job is done in 10.0 s. What
Solution:
Pg264pp9
power
is developed
in watts and kilowatts?
F = 575N, d = 20.0m, t = 10.0s, P = ?
W  Fd  575 N  20.0m  1.15 104 J
W 1.15 104 J
P

 1.15 103W
t
10.0s
 kW 
 1.15 10 W  
  1.15kW
 1000W 
3
19
Practice:
Does the centripetal force acting on an object do
work on the object?
a) Yes, since a force acts and the object moves,
and work is force times distance.
b) Yes, since it takes energy to turn an object.
c) No, because the object has constant speed.
d) No, because the force and the velocity of the
object are perpendicular.
20
Practice: Pg264pp12
An electric motor develops 65 kW of power as it
lifts a loaded elevator 17.5 m in 35.0 s. How
much force does the motor exert?
P = 65kW = 65000 W, d = 17.5 m, t = 35.0 s
F=?
d 17.5m
m


0.500
v
t 35.0s
s
P  Fv  F 
P
21
P 65000W

 1.3 105 N
v 0.500 m
s
W
6
 W  Pt  65000W  35.0s  2.275 10 J
t
W 2.275 106 J

 1.3 105 N
W  Fd  F 
d
17.5m
Kinetic Energy, KE
KE 
Energy or ability to
do work because of
motion.
1 2
mv
2
 m: mass
 v: velocity
Unit:
2
m
m2 
m
kg

 m  N m  J
kg


kg


m
v




 KE  

 
2 
2
s 
s

s
2
22
Practice:
Car J moves twice as fast as car K, and car J has
half the mass of car K. The kinetic energy of
car J, compared to car K is
a) The same.
b) 2 to 1.
c) 4 to 1.
d) 1 to 2.
23
Work-Kinetic Energy Theorem
Work and energy are related by
Wnet  KE  KE f  KEi


24
Wnet: total work done on an object
KE: kinetic energy of that object
Example
You push a 10-kg desk, initially at rest, with a force of 100
N a distance of 8.5 m across the classroom. What is the
final speed of the desk?
m  10kg , F  100 N , d  8.5m, vi  0, v f  ?
W  KE  KE f  KEi
1
1
2
2
Fd  mv f  mvi
2
2
2Fd
1
2
2
Fd  mv f  v f 
m
2
vf 
25
2 Fd
2 100 N  8.5m
m

 13
m
10kg
s
Practice:
A rock of mass 2.0 kg is thrown with a speed of 10 m/s at
an angle of 40o above the horizontal at the edge of a 30m high cliff. With what speed does it hit the ground?
m
m  2.0kg , i  40 , h  30m, vi  10 , v f  ?
s
W  KE  KE f  KEi
1
1
mg h  mv f 2  mvi 2
2
2
o
2gh  v f 2  vi 2
v f 2  vi 2  2 g h
v f  vi  2 g h 
2
26
2
m
m
 m
10

2

9.8


30
m

26




2
s
s
s


Practice:
If the net work done on an object is positive, then
the object’s kinetic energy
a) decreases.
b) remains the same.
c) increases.
d) is zero.
27
Practice:
An arrow of mass 0.020 kg is shot horizontally
into a bale of hay, striking the hay with a
velocity of 60 m/s. It penetrates a depth of
0.20 m before stopping. What is the average
stopping force acting on the arrow?
a) 45 N
b) 90 N
c) 180 N
d) 360 N
28
Simple Machine



Simple Machine: anything that can change the
force (magnitude or direction) to accomplish a
task
Does not change the amount of work done.
(Then why use a machine?)
Example:
–
Lever, pulley, wheel-and-axle,
inclined plane, wedge, and screw
29
Mechanical Advantage, MA
Fo
Fr
MA 

Fi
Fe

Fi or Fe: input force or effort force
–
–

Fo or Fr: output force or resistance force
–
–

30
Force exerted by the machine on the load.
Force you need to exert if not using a machine.
MA has no unit.
–

Force you actually exert on a machine
Force you actually exert when using a machine
MA has a unit of 1, or MA is dimensionless.
MA > 1: machine increases your force
Input and Output Work
Wi  Fi di



Wi: input work, work done by you when using machine
Fi: input (effort) force
di: input (effort) displacement, displacement of your hand
Wo  Fo d o



31
Wo: output work, work given out by machine. (Work you
need to do if not using the machine.)
Fo: output (resistance) force
do: output (resistance) displacement, displacement caused
by machine, displacement when not using machine
Ideal Machine
Ideal Machine: Wi  Wo
Ideal Mechanical Advantage: IMA  MAI 
MA 
Fo
is valid for ideal and non-ideal.
Fi
IMA 
32
di
is valid only for ideal machine.
do
di
do
Efficiency, eff
Efficiency: ratio of the work done by the machine to the
work put into the machine; how much of the input work
converted into output work by the machine
Eff 
Also
33
Wo
 100%
Wi
MA
Eff 
100%
IMA
Efficiency has no unit.
Example: Pg272pp25
A sledge hammer is used to drive a wedge into a log to split it.
When the wedge is driven 0.20 m into the log, the log is
separated a distance of 5.0 cm. A force of 1.7  104 N is
needed to split the log, and the sledge exerts a force of 1.1 
104 N .
a) What is the IMA of the wedge?
b) Find the MA of the wedge.
c) Calculate the efficiency of the wedge as a machine.
di = 0.20m, do = 5.0cm = 0.05m, Fo = 1.7  104 N, Fi = 1.1  104 N
a ) IMA  ?
d
0.20m
 4.0
IMA  i 
d o 0.05m
c)eff  ?
b) MA  ?
Fo 1.7 10 N

 1.5
MA 
4
Fi 1.110 N
4
34
1.5
MA

100%  38%

100%
eff 
4.0
IMA
Practice: Pg272pp26
A worker uses a pulley system to raise a 24.0-kg carton 16.5 m.
A force of 129 N is exerted and the rope is pulled 33.0 m.
a) What is the MA of the pulley system?
b) What is the efficiency of the system?
Fo  mg  24kg  9.8m / s 2  235N , d o  16.5m, Fi  129 N , d i  33.0m
a) MA  ?
Fo 235 N

 1.82
MA 
Fi 129 N
b)eff  ?
Wo  Fo do  235N 16.5m  3878J
Wi  Fi di  129 N  33.0m  4257 J
35
Wo 3878 J

 0.911  91.1%
eff 
Wi 4257 J
di 33.0m

 2.00,
IMA 
d o 16.5m
MA
 100%
Eff 
IMA

1.82
100%  91%
2.00
Inclined Plane
di
IMA 
do
d
1
IMA  
h sin 
36
d

h