Transcript Lec 7

AP Physics Chapter 7
Kinetic Energy and
Work
1
AP Physics



2
Turn in Chapter 6 Homework, Worksheet &
Lab Report
Lecture
Q&A
Kinetic Energy, K

Kinetic Energy: ability to do work due to motion
1 2
K  mv
2




3
–
–
m: mass of object
v: velocity of object
Unit: [K] = [m] · [v]2 = kg · (m/s)2 = Joule = J
1 electron-volt = 1 eV = 1.60  10-19 J
Scalar, no direction
K0
Example:
What is the kinetic energy associated with the Earth’s
orbiting of the Sun? (Mass of Earth is 5.98  1024 kg, and
radius of Earth orbiting around sun is 1.50  1011 m.)
m = 5.98  1024 kg, T = 365 dys = 3.15  107 s
r = 1.50  1011 m, K = ?
2 r
v

T
2  1.50 1011 m 
3.15 107 s
m
 2.99 10
s
4
2
1 2 1

24
4 m
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K  mv    5.98 10 kg    2.99 10

2.67

10
J

2
2
s

4
Work, W

Work: energy transferred to or from an object
by means of a force acting on the object

Work done by object A (to object B)


5
W > 0 if energy transferred (from A) to object B
W < 0 if energy transferred from object B (to A)
F

d
Work Done by A Constant Force
W  F  d  Fd cos   Fx d x  Fy d y  Fz d z




W: work done by force on the object
F: constant force acting on object
d: displacement of object with force acting on, x
: angle between F and d




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
 < 90o  W > 0
 > 90o  W < 0
 = 90o  W = 0
W    F d 
W: a scalar, no direction (but W can be negative)
Valid for particle-like objects
Work Done by Constant Forces




W   F  d  F 1  F 2  F 3  ...  d
 F1  d  F 2  d  F 3  d  ...
 W1  W2  W3  ...


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Find the net force (vectors), then find the dot
product of the net force and the displacement,
or
Find the work done by each individual force, then
add up all the works done by each force.
Practice: Pg160-8
A floating ice block is pushed through a displacement d = (15 m)i –
(12 m) j along a straight embankment by rushing water, which exerts
a force F = (210 N)i – (150N )j on the block. How much work does
the force do on the block during the displacement?


ˆ
ˆ
d  15mi  12m j, F  210N iˆ  150N  ˆj,
W ?
W  F d
 Fx dx  Fy d y
  210 N   15m    150 N    12m 
 3200 J  1800 J
 5000 J  5.0 103 J  5.0kJ
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Work-Kinetic Energy Theorem

The total work done on an object is equal to
the change in the kinetic energy of that object
W  K  K f  Ki

Another look at work:
  

W  F  d  F  vt  F  v t

If F and d are in same direction:
W  Fd
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Practice:
A single force acts on a body that moves along a straight line. The
diagram shows the velocity versus time for the body. Find the sign
(plus or minus) of the work done by the force on the body in each of
the intervals AB, BC, CD, and DE.
v
W = K
AB:
BC:
10
K: __
0
__
+  K __
> 0
 WAB __
> 0
v __
= 0  K __
= 0
 WBC __
= 0
B
C
+
A
-
CD:
K: __
+  __
0  K __
< 0
 WCD __
< 0
DE:
K: __
0  __
+ (though v: __
0  __)
-  WDE __
> 0
D
t
E
Practice: Pg160-13
Figure 7-29 shows three forces applied to a greased trunk that moves
leftward by 3.00 m over a frictionless floor. The force magnitudes are
F1 = 5.00 N, F2 = 9.00 N, and F3 = 3.00 N. During the displacement,
(a) what is the net work done on the trunk by the three forces and (b)
does the kinetic energy of the trunk increase or decrease?
F1  5.00 N , 1  0o , F2  9.00 N , 2  120o ,
F3  3.00 N , 3  90o , d  3.00m,
a ) W  ?
F2
60o
F1
W  Fd cos 
W1  5.00N  3.00m  cos0o  15.0J
W2  9.00N  3.00m  cos120o  13.5J
W3  3.00N  3.00m  cos90o  0J
W  W1  W2  W3  15.0J  13.5J  0  1.5J
b)K   W  0  K increases
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F3
W  Fd cos 
Work (Wg) Done by Weight (Fg)
Wg  mgd cos 




m: mass
g = 9.8 m/s2
d = displacement
: angle between d and downward direction
h
180o
Wg  mg h

12
mg
h: change in object’s height
?

Going up  h > 0  Wg < 0  Weight doing negative work

Going down  h __
______ work
< 0  Wg ___
> 0  Weight doing positive
Practice: Pg164-59
To push a 25.0 kg crate up a frictionless incline, angled at 25.0o to the
horizontal, a worker exerts a force of 209 N, parallel to the incline. As
the crate slides 1.50 m,
a) How much work is done on the crate by the worker’s applied force,
b) How much work is done on the crate by the weight of the crate, and
c) How much work is done on the crate by the normal force exerted by
the incline on the crate?
d) What is the total, work done on the crate?
m  25.0kg,  25.0o , F  209N , d  1.50m
a)  0o ,Wapp  ?
Wapp  Fd cos   209N 1.50m  cos0o  314J
d

b)Wg  ?
s2
g
W  mg h  mgd sin   25.0kg  9.8 m 1.50m  sin 25.0o  155 J
c)WN  ?
  90o  WN  0
d ) W  ?
 W  Wapp  Wg  WN  314 J  155J  159 J
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h
Practice: Pg161-19
A cord is used to vertically lower an initially stationary
block of mass M at a constant downward acceleration of
g/4. when the block has been lowered a distance d, find
a) the work done by the cord’s force on the block,
b) the work done by the weight of the block,
c) the kinetic energy of the block, and
d) the speed of the block.
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Solution: Pg 161-19
Let downward be positive direction, then
 F  W  T  ma  mg / 4
 T  W  mg / 4  mg  mg / 4  3mg / 4
a)  180o ,WT  ?
WT  Td cos    3mg / 4  d cos180o  3mgd / 4
T
b)  0o ,Wg  ?
Wg  mgd cos 0o  mgd
+
c) K f  ?
W
K  K f  Ki   W  WT  Wg  3/ 4mgd  mgd  mgd / 4
 K f  Ki  mgd / 4  mgd / 4
d )v  ?
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2  mgd / 4 
1 2
2K
v



K  mv 
m
m
2
gd
2
Work Done by a Variable Force
xf
W  Fd  W   F ( x)dx
1-D
xi
xf
W 
yf
 F ( x)dx   F ( y )dy   F ( z )dz
xi
yi
or: Area under curve of
Force-position graph
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zf
zi
3-D
Example: Pg162-31
A 10 kg mass moves along an x axis. Its acceleration as a
function of its position is shown in Fig. 7-37. What is the
net work performed on the mass by the force causing the
acceleration as the mass moves from x = 0 to x = 8.0 m?
m  10kg, xi  0, x f  8m,W  ?
xf
W
 F ( x)dx
xi
xf

 ma( x)dx 
xi
xf
m  a ( x )dx
xi
 m   Area Under Curve
m
1
 10kg    8m  20 2   800 J
s 
2
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Spring Force


When a spring is
compressed or
stretched, it exerts a
force on the object
that compresses or
stretches it.
Force is opposite to
change of length of
spring
 Restoring force
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F
x
F
x
Hooke’s Law

Spring force is given by F  k x




or
F  kx
F: Force exerts by spring, not by you
k: Spring constant, property of spring, stiffness of
spring (larger k, stiffer spring)
x: Displacement of (end of) spring from equilibrium
position, x
-: F and x are in opposite direction
Notice that this force is the force exerted by the spring, not
the force you apply to compress or stretch the spring.
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Ideal (Perfect) Spring




20
No mass
Can always go back to original length when not
compressed or stretched.
Does not become warm/hot
x is small compared to the entire length of
spring
Work Done by Spring (Force)
Ws  x Fdx  x  kx  dx 
xf
xf
i
i
1 2 1
kxi  kx f 2
2
2

This is the work done by the spring, not the work done
(by you) to compress or stretch the spring.

If the spring is initially at equilibrium position, then xi = 0,
1 2 1
1
and
2
2
Ws 
kxi  kx f   kx f  0
2
2
2
F
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Ws 
xf
x
- work ?
x
Xi = 0
F
xf
Example:
A spring with a spring constant of 15 N/cm has a cage attached to one end.
How much work does the spring force do on the cage if the spring is
stretched from its relaxed length by 7.6 mm?
How much additional work is done by the spring force if it is stretched by an
additional 7.6 mm?
equilibrium
7.6 mm
k  15
7.6 mm
N 100cm
N

 1500
cm
m
m
a) xi  0, x f  7.6mm  7.6 103 m,Ws  ?
2
1 
N
1 2 1
3
2


1500

7.6

10
m
 0.043J
Ws  kxi  kx f 


 
2 
m
2
2
b) xi  7.6 103 m, x f  2  7.6mm  1.52 102 m,Ws  ?
1 2 1
1
kxi  kx f 2  k  xi 2  x f 2 
2
2
2
2
2
1 
N 
3
2
  1500    7.6  10 m   1.52  10 m    0.13J

2 
m 
Ws 
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Practice:
The only force acting on a 2.0 kg body as it
moves along the positive x axis has an x
component Fx = -6x N, where x is in meters.
The velocity of the body at x = 3.0 m is 8.0
m/s.
a) What is the velocity of the body at x = 4.0 m?
b) At what positive value of x will the body have
a velocity of 5.0 m/s?
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m  2.0kg, k  6N / m, xi  3.0m, vi  8.0m / s
a) x f  4.0m, v f  ?
W
Solution
K 

xf
xi
Fdx  
4.0
3.0
 6x  dx   3x  3.0 
2
4.0
1
1
1
2W
mv f 2  mvi 2  m  v f 2  vi 2   W  v f 2 
 vi 2
2
2
2
m
 vf 
2W
 vi 2 
m
2  21J  
m
m
  8.0   6.6
2.0kg
s
s

2
b)v f  5.0m / s, x f  ?
2
2


1
1
m
m




2
2
W  K  m  v f  vi    2kg   5.0    8.0    39 J
2
2
s 
s  

W
24

xf
xi
 xf 
Fdx 
3.0  6 x  dx   3x  3.0  3x  x
xf
2
xf
27   39
27  W

 4.7m
3
3
2
3.0
f
 27  3x f 2
21J
Power


Power: the (time) rate at which work is done by a force,
or the (time) rate at which energy is transferred (from
one object to another)
Average Power:
W
P
–
–

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t
W: work (scalar, no direction)
t: time, duration, for that amount of work done
Unit of Power:
W  
 P  t

J
 Watt  W ;
s
1 horsepower = 1 hp = 746 W
Instantaneous Power
d  F  x
dx
dW
F
 F v

P
dt
dt
dt
 Fv cos  or Fx vx  Fy v y  Fz vz
F: force, vector
 v: velocity, vector
 : angle between F and v

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Example: Pg165-73
An elevator cab has a mass of 4500 kg and can carry a
maximum load of 1800 kg. If the cab is moving upward at
full load of 3.80 m/s, what power is required of the force
moving the cab to maintain that speed?
M  m1  m2  4500kg  1800kg  6300kg ,
m
v  3.80 , P  ?
s
m
F  Mg  6300kg  9.8 2  61700 N
s
 0
m
P  Fv cos   61700 N  3.80  2.34  105W
s
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Practice: Pg162-44
a)
b)
28
At a certain instant, a particle experiences a force F =
(4.0 N)i – (2.0 N)j + (9.0 N)k while having a velocity v
= -(2.0 m/s)i + (4.0 m/s)k. What is the instantaneous
rate at which the force does work on the particle?
At some other time, the velocity consists of only a j
component. If the force is unchanged, and the
instantaneous power is –12 W, what is the velocity of
the particle just then?
Solution: Pg162-44
m 
m

a) F   4.0 N  iˆ   2.0 N  ˆj   9.0 N  kˆ, v   2.0  iˆ   4.0  kˆ, P  ?
s 
s

P  F  v  Fx vx  Fy v y  Fz vz
m
m


  4.0 N   2.0    2.0 N  0    9.0 N   4.0   28W
s
s


b) v = ?
Let the velocity be v = vy j, then
P  F  v  Fx v x  Fy v y  Fz v z   2.0 N   v y 
vy 
29
P
12W
m

 6.0
2.0 N 2.0 N
s
So, v = (6.0m/s) j
Practice: Pg162-45
A fully loaded, slow-moving freight
elevator has a cab with a total mass
of 1200 kg, which is required to
travel upward 54 m in 3.0 min. The
elevator’s counterweight has a
mass of only 950 kg. so, the
elevator motor must help pull the
cab upward. What average power
(in horsepower) is required of the
force that motor exerts on the cab
via the cable?
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Solution: Pg162-45
me  1200kg , mc  950kg , h  54m, t  3min
P?
Three things doing work:
Work done by gravity on elevator cab:
We = -meg h = -1200kg  9.8m/s2  54 m = -6.35  105 J
Work done by gravity on counterweight:
Wc = -mcgh = -950kg  9.8m/s2  (-54 m) = 5.03  105 J
Work done by motor:
Wm = ?
Constant velocity: W = 0  Wm + We + Wc = 0
So, Wm = -(We + Wc ) = - (-6.35  105 J + 5.03  105 J) = 1.32  105 J
Then average power is
31
Wm 1.32  105 J min
 hp 
P


 733W  
  0.98hp
t
3.0 min
60s
 746W 
Another approach
The motor essentially needs to do work to raise the cab (1200 kg) 54
m and lower the counterweight (950kg) 54 m. So equivalently, the
motor needs to do work to raise a mass of 250 kg (1200kg –
950kg) 54 m.
The force needed is
F  mg  250kg  9.8
m
 2450 N
2
s
So the work done by the motor is
W  Fd  2450 N  54m  132300 J
And the power is
P
32
W 132300 J

 735W  0.98hP
t
180s
Reference Frames

Invariant quantities: quantities measured exactly
the same at different inertial reference frames
–

Variant quantities:
–

force, mass, acceleration, time
displacement, velocity
Principle of Invariance:
The laws of physics must have the same form in
all inertial reference frames.
33