Transcript 27.11.2012 - Erwin Sitompul

Lecture 12
Ch7. Work-Kinetic Energy Theorem
University Physics: Mechanics
Dr.-Ing. Erwin Sitompul
http://zitompul.wordpress.com
2012
Energy
 Energy is a number associated with the state (or condition) of
a system of one or more objects.
 If a force changes one of the objects by making it move, then
the energy number is said to be change.
 After countless experiments, scientists and engineers realized
that energy numbers can be assigned to or withdrawn from a
system by conducting a certain scheme.
 If planned carefully, the
outcomes of the experiments
can be predicted and made
useful. For example: flying
machine, refrigerator, etc.
Erwin Sitompul
University Physics: Mechanics
12/2
Conservation of Energy
 Energy can be transformed from one type to another and
transferred from one object to another, but the total amount is
always the same. Thus, energy is conserved.
Erwin Sitompul
University Physics: Mechanics
12/3
Kinetic Energy
 Kinetic energy K is energy associated with the state of
motion of an object.
 The faster the object moves, the greater is its kinetic energy.
When the object is stationary, its kinetic energy is zero.
 For an object of mass m whose speed v is well below the
speed of light,
K 
1
2
mv
2
(kinetic energy)
 The SI unit of kinetic energy (and every other type of energy)
is joule (J), defined as:
1 joule  1 J  1 kg  m
Erwin Sitompul
2
s
2
University Physics: Mechanics
12/4
Example: Locomotives Crash
In 1896 in Texas, two locomotives were deliberately crashed
frontally to each other for a publicity stunt in front of 40,000
spectators. They are parked at opposite ends of a 6.4-km-long
track before fired up, and crashed at full speed. Hundreds of
people were hurt by flying debris, several were killed.
Assuming each locomotive weighed 1.2×106 N and its
acceleration was a constant 0.26 m/s2, what was the total
kinetic energy of the two locomotives just before the collision?
Erwin Sitompul
University Physics: Mechanics
12/5
Example: Locomotives Crash
a  0.26 m s
6400 m
2
a   0.26 m s
2
v  v0  2 a ( x  x0 )
2
2
v  0  2(0.26)(3200)
2
v   4 0 .7 9 m s
m 
W
g
1.2  10 N
6

9.8 m s
2
 1.224  10 kg
5
1
8
5
2
2 
K  2  m v   (1.224  10 )(40.79)  2 .0 3 7  1 0 J
2

Erwin Sitompul
University Physics: Mechanics
12/6
Work
 Work W is energy transferred to or from an object by means
of a force acting on the object.
 Energy transferred to the object is positive work, and energy
transferred from the object is negative work.
 Work has the same unit as energy, Joule, and is a scalar
quantity.
Erwin Sitompul
University Physics: Mechanics
12/7
Work and Kinetic Energy
 The work a force does on an object can be calculated as the
object moves through some displacement.
 Only the force component along the object’s displacement will
be used.
 The force component perpendicular to the displacement
does zero work.
W  F d cos 
(work done by a constant force)
 We can also write in scalar (dot) product as:
W  F d
Erwin Sitompul
(work done by a constant force)
University Physics: Mechanics
12/8
Work and Kinetic Energy
 Assumptions must be taken to use these formulas to calculate
work done on an object by a force.
 The force must be a constant force, not change in
magnitude or direction
 The object must be particle-like and rigid
 The SI unit of work is joule (J), defined as:
1 joule  1 J  1 kg  m
Erwin Sitompul
2
s 1 N m
2
University Physics: Mechanics
12/9
Flashback: Multiplying Vectors
The Scalar Product
→
→
→→
 The scalar product of the vector a and b is written as a·b
and defined to be:
a  b  ab cos 
→→
 Because of the notation, a·b is also known as the dot
product and is spoken as “a dot b.”
→
→
 If a is perpendicular to b, means Φ = 90°, then the
dot product is equal to zero.
→
→
 If a is parallel to b, means Φ = 0, then the dot product
is equal to ab.
Erwin Sitompul
University Physics: Mechanics 12/10
Flashback: Multiplying Vectors
 The dot product can be regarded as the product of the
magnitude of the first vector and the projection magnitude
of the second vector on the first vector
a  b  ab cos 
 ( a cos  )( b )
 ( a )( b cos  )
Erwin Sitompul
University Physics: Mechanics 12/11
Flashback: Multiplying Vectors
 When two vectors are in unit vector notation, their dot
product can be written as
ˆ  ( b ˆi  b ˆj  b k)
ˆ
a  b  ( a x ˆi  a y ˆj  a z k)
x
y
z
 a x bx  a y b y  a z bz

ˆi
ˆj
kˆ
Erwin Sitompul
ˆi
ˆj
kˆ
1
0
0
0
1
0
0
0
1
University Physics: Mechanics 12/12
Flashback: Multiplying Vectors
→
^
→
^
^
^
What is the angle Φ between a = 3i – 4j and b = –2i + 3k ?
z
Solution:
3
a  b  ab cos 
a
3  (  4)  5
b
(  2)  3  3.606
2
2
2
–4
–2
y
a
2
a  b  (3iˆ  4 ˆj)  (  2 ˆi  3kˆ )
ˆ  2 ˆi)
 (3i)(
 6

b
3
x

ˆi
ˆj
kˆ
ˆi
ˆj
kˆ
1
0
0
0
1
0
0
0
1
 6  (5)(3.606) cos 
6
1
 109.438 
  cos
(5)(3.606)
Erwin Sitompul
University Physics: Mechanics 12/13
Example: Calculating Work
→
Calculate the work done by→F along the
direction of displacement Δs, as shown on
the next figure.
W  F  s
 (3iˆ  4 ˆj)  (2 ˆi  5 ˆj)
 (3)(2)  (4)(  5)
  14 N  m
Erwin Sitompul
University Physics: Mechanics 12/14
Work-Kinetic Energy Theorem

 
change in the kinetic
the w ork done on

energy of a particle
the article

K  K f  K i  W

 
 
kinetic energy after
kinetic energy
the ne t


the net w ork is done
before the net w ork
w ork done
K
f

 Ki W
 These statements are known traditionally as the work-kinetic
energy theorem for particles.
 They hold for both positive and negative work. If the net work
done on a particle is positive, then the particle’s kinetic energy
increases by the amount of that work. And so as the opposite
case.
Erwin Sitompul
University Physics: Mechanics 12/15
Example: Industrial Spies
Two industrial spies are
→ sliding an initially stationary 225 kg floor
safe a displacement d of magnitude 8.5 m, straight toward their
truck.
→
The push F1 of spy 001 is 12 N, directed→ at an angle of 30°
downward from the horizontal; the pull F2 of spy 002 is 10 N,
directed at 40° above the horizontal.
Erwin Sitompul
University Physics: Mechanics 12/16
Example: Industrial Spies
(a) What is the net work done
→ on the safe by forces
during the displacement d?
→
F1
and
→
F2
W1  F1 d cos 1
 (12)(8.5) cos(  30  )
 88.33 J
W 2  F2 d cos  2
 (10)(8.5) cos(40  )
F1 = 12 N, F2 = 10 N
 65.11 J
W  W1  W 2
 88.33  65.11
 153.44 J
Erwin Sitompul
University Physics: Mechanics 12/17
Example: Industrial Spies
(b) During the displacement, what →is the work Wg done on the
safe by the gravitational force Fg and what
is the work WN
→
done on the safe by the normal force FN from the floor?
• The forces are perpendicular to the
displacement of the safe
• They do zero work on the safe and do
not transfer any energy to or from it.
Erwin Sitompul
University Physics: Mechanics 12/18
Example: Industrial Spies
(c) The safe is initially stationary. What is its speed vf at the
end of the 8.5 m displacement?
K  K f  K i  W ,
1
mv  W
2
Erwin Sitompul
2
 v
Ki  0
2W
m

2(153.44)
 1.168 m s
(225)
University Physics: Mechanics 12/19
Example: Crate of Crepe
During a storm, a crate of crepe is sliding
across a slick, oily
→
parking lot through a displacement d = –3 ^i →m while a steady
wind pushes against the crate with a force F = 2 ^i – 6 ^j N. The
situation and coordinate axes are shown on the next figure.
(a) How much work does this force do on the crate during the
displacement?
W  F d
 (2 ˆi  6 ˆj) N  (  3iˆ ) m
 6 N  m
 6 J
Erwin Sitompul
• The force does a negative 6 J
of work on the crate
• Other way stated, the force
transfers 6 J of energy out
from the kinetic energy of the
crate
University Physics: Mechanics 12/20
Example: Crate of Crepe
During a storm, a crate of crepe is sliding
across a slick, oily
→
^i while a steady
parking lot through a displacement d = –3 m
→
wind pushes against the crate with a force F = 2 ^i – 6 ^j N. The
situation and coordinate axes are shown on the next figure.
(b) If the crate has a kinetic energy of 10 J at the beginning
of displacement d, what is its kinetic energy at the end of
d?
K
f
 Ki W
 (10)  (  6)
4J
Erwin Sitompul
University Physics: Mechanics 12/21
Work Done by the Gravitational Force
 The figure shows a particle-like tomato of
mass m that is thrown upward with initial
speed v0 and thus with initial kinetic energy
Ki = 1/2mv02.
 As the tomato rises,
→ it is slowed by a
gravitational force Fg; that is, the tomato’s
→
kinetic energy decreases because Fg does
work on the tomato as it rises.
→
 The work Wg during displacement
d done by
→
the gravitational force Fg is
W g  m gd cos 
(work done by gravitational force)
W g  m gd cos 180    m gd
• During the object’s rise, the gravitational force acting
on the object transfer energy in the amount mgd
from the kinetic energy of the object.
• What happens when the tomato falls back down?
Erwin Sitompul
University Physics: Mechanics 12/22
Work Done in Lifting and Lowering an Object
 Now suppose we lift a particle-like
object
by applying a vertical force
→
F to it.
 During the upward displacement,
our applied force does positive
work Wa on the object while the
gravitational force does negative
work Wg on it.
K  K
f
 K i  Wa  Wg
 In one common situation, the
object is stationary before and
after the lift, that means
Wa  Wg  0
W a  W g
W a   m gd cos 
Erwin Sitompul
• Raising
W a   m gd
• Lowering
W a   m gd
• Raising,  = 180°, Wa > 0,
work is exerted
• Lowering,  = 0°, Wa < 0,
work is received
University Physics: Mechanics 12/23
Work Done in Lifting and Lowering an Object
An initially stationary 15.0 kg crate of cheese wheels is pulled,
via a cable, a distance d = 5.70 m up a frictionless ramp to a
height h of 2.50 m, where it stops.
(a) How much work Wg is done on the crate by the gravitational
force Fg during the lift?
W g  m gd cos 
 m gd cos(90    )
  m gd sin 
Erwin Sitompul
  (1 5)(9 .8)(5 .7 )  52 .7.5 
  367.5 J
University Physics: Mechanics 12/24
Work Done in Lifting and Lowering an Object
An initially stationary 15.0 kg crate of cheese wheels is pulled,
via a cable, a distance d = 5.70 m up a frictionless ramp to a
height h of 2.50 m, where it stops.
(b) How much work WT is done on the crate by the force T from
the cable during the lift?
W T  W g
  367.5 J
Erwin Sitompul
University Physics: Mechanics 12/25
Work Done in Lifting and Lowering an Object
An elevator cab of mass m = 500 kg is descending
with speed vi = 4.0 m/s when its supporting cable
begins to slip,→allowing
it to fall with constant
→
acceleration a = g/5.
(a) During the fall through a distance d = 12 m,
what is the work Wg done on the cab by the
gravitational force Fg?
W g  m gd cos 0   (500)(9.8)(12)  58800 J
(b) During the 12 m fall, what is the work WT done on
the cab by the upward pull T of the elevator cable?
T  Fg  m a  T  Fg  m a
W T  Td cos 
 ( F g  m a ) d cos 180 
  (5 0 0 )(9 .8  9 .8 5)(1 2 )
  47040 J
  m ( g  a )d
Erwin Sitompul
University Physics: Mechanics 12/26
Work Done in Lifting and Lowering an Object
An elevator cab of mass m = 500 kg is descending
with speed vi = 4.0 m/s when its supporting cable
begins to slip,→allowing
it to fall with constant
→
acceleration a = g/5.
(c) What is the net work W done on the cab
during the fall?
W  W T  W g  58800 J  47040 J  11760 J
(d) What is the cab’s kinetic energy at the end of the
12 m fall?
K
f
 Ki W

1
2
m vi  W

1
2
(500)(4)  11760
2
2
 15760 J
Erwin Sitompul
University Physics: Mechanics 12/27
Exercise Problems
1. A skier with a mass of 58 kg slides downhill a
surface with 25° inclination. Her initial velocity
is 3.6 m/s. If she experiences a friction force
of 70 N, determine her velocity 57 m further
by using work-kinetic energy theorem.
Answer: vf = 18.64 m/s.
2. An 8.0-kg object is moving in the positive
direction of an x axis. When it passes through
x = 0, a constant force directed along the axis
begins to act on it. The next figure gives its
kinetic energy K versus position x as it moves
from x = 0 to x = 5 m. K0 = 30 J.
The forces continues to act. What is v when
the object moves back through x = –3 m?
Hint: Use work-kinetic energy theorem.
Answer: 3.46 m/s.
Erwin Sitompul
University Physics: Mechanics 12/28
End of the Lecture
Erwin Sitompul
University Physics: Mechanics 12/29