Transcript Work

Section 7.5
Scientific and Statistical Applications
Review: Hooke’s Law:
F  kx
A spring has a natural length of 1 m.
A force of 24 N stretches the spring to 1.8 m.
F  kx
24  k .8
a Find k:
30  k
F  30x
b How much work would be needed to stretch the spring
3m beyond its natural length?
W   F  x  dx
b
a
3
W   30 x dx
0
W  15x
2 3
0
W  135 newton-meters
Remember when work is constant,
Work = Force ● Displacement
Remember when work is not constant,
b
Work =
 F ( x)dx
a
Book Ex. 2
A leaky 5 lb bucket is raised 20 feet
The rope weighs 0.08 lb/ft.
20
The bucket starts with 2 gal (16 lb) of water
and is empty when it just reaches the top.
Work:
Bucket (constant): 5 lb  20 ft  100 ft-lb
Water: The force is proportional to
remaining rope.
0
4
20  x
F  x 
16  16  x
5
20
W   F  x  dx
b
a
A leaky 5 lb bucket is raised 20 feet
The rope weighs 0.08 lb/ft.
20
The bucket starts with 2 gal (16 lb) of water
and is empty when it just reaches the top.
Work:
Bucket: 5 lb  20 ft  100 ft-lb
Water:
0
4
W   16  x dx
0
5
20
2 2
W  16 x  x
5 0
20
2  202
 160 ft-lb
W  16  20 
5
A leaky 5 lb bucket is raised 20 feet
The rope weighs 0.08 lb/ft.
20
The bucket starts with 2 gal (16 lb) of water
and is empty when it just reaches the top.
Work:
Bucket: 5 lb  20 ft  100 ft-lb
W  160 ft-lb
Water:
Rope: F  x    20  x  0.08
0
W 
20
0
1.6  .08x  dx
W  1.6 x  .04 x
2 20
0
 16 ft-lb
Total: 100  160  16  276 ft-lb
Like Ex. 3
5 ft
4 ft
I want to pump the water out of this
tank. How much work is done?
w  Fd
The force is the weight of the water.
The water at the bottom of the tank
must be moved further than the
water at the top.
10 ft
4 ft
0
Consider the work to move one “slab”
of water:
weight of slab  density  volume
 62.5    52 dx
10 ft
 1562.5 dx
dx
10
5 ft
distance  x  4
5 ft
4 ft
I want to pump the water out of this
tank. How much work is done?
w  Fd
weight of slab  density  volume
 62.5    52 dx
10 ft
 1562.5 dx
4 ft
0
distance  x  4
work   x  41562.5 dx
distance
10 ft
dx
10
5 ft
W 
10
0
force
 x  41562.5
dx
5 ft
4 ft
I want to pump the water out of this
tank. How much work is done?
work   x  41562.5 dx
10 ft
distance
W 
10
0
4 ft
0
A 1 horsepower pump,
rated at 550 ft-lb/sec,
could empty the tank
10 ft
in just under 14 dx
minutes!
10
5 ft
force
 x  41562.5
dx
10
1 2

W  1562.5  x  4 x 
2
0
W  1562.5 50  40
W  441, 786 ft-lb
Example 3 (book)
A conical tank is filled to within 2 ft of
2 ft
the top with salad oil weighing 57 lb/ft3.
10 ft
How much work is required to pump
the oil to the rim?
10 ft
Consider one slice (slab) first:
W y   F  d
10  y
5,10 W  density  volume  distance
 y
x
y  2x
y
x
1
y
2
W y 
  1 2 
 57   y   dy10  y 
  2  
1 2
W   10  y  57  y dy
0
4
8
Example 3 (book)
A conical tank if filled to within 2 ft of
2 ft
the top with salad oil weighing 57 lb/ft3.
10 ft
How much work is required to pump
the oil to the rim?
10 ft
W y 
5,10
10  y
x
y  2x
y
x
1
y
2
  1 2 
 57   y   dy10  y 
  2  
1 2
W   10  y  57  y dy
0
4
57 8
2
3
W
10 y  y dy

4 0
4 8
57 10 3 y 
W
 y  
4 3
4 0
8
Example 3 (book)
A conical tank if filled to within 2 ft of
2 ft
the top with salad oil weighing 57 lb/ft3.
10 ft
How much work is required to pump
the oil to the rim?
10 ft
5,10
10  y
x
y  2x
y
x
1
y
2
57
W
4

57
W
4
10 3 y 
 y  
4 0
3
8
0
10 y 2  y 3 dy
4
8
57  5120 4096 
W


4  3
4 
W  30,561 ft-lb
What is the force on the bottom of
the aquarium?
2 ft
1 ft
3 ft
Force  weight of water
 density  volume
lb
 62.5 3  2 ft  3 ft 1 ft
ft
 375 lb
What is the force on the front face
of the aquarium?
2 ft
Depth (and pressure) are not constant.
3 ft
1 ft
0
If we consider a very thin horizontal
strip, the depth doesn’t change much,
and neither does the pressure.
Fy  62.5  y  3 dy
depth
density
area
y
2
2 ft
dy
2
3 ft
F   62.5  y  3 dy
0
2
It is just a
coincidence
that this
matches the
first answer!
187.5 2
F
y  375 lb
2
0
A flat plate is submerged vertically
We could have put
as shown. (It is a window in the
the origin at the
surface, but the math shark pool at the city aquarium.)
was easier this way.
2 ft
Find the force on one side of the
plate.
yx
x y
3 ft
6 ft
Depth of strip:  5  y 
Length of strip: 2 x  2 y
Area of strip: 2 y dy
3
Fy  62.5  5  y  2 y dy
F  125 5 y  y 2 dy
density depth area
5 2 1 3
F  125  y  y 
3 0
2
0
3
F   62.5  5  y  2 y dy
3
0
F  1687.5 lb
Normal Distribution:
13.5%
34%
3 2  
2.35%
 2 3
68%
95%
99.7%
“68, 95, 99.7 rule”
For many real-life events, a
frequency distribution plot
appears in the shape of a
“normal curve”.
Examples:
The mean  (or x ) is in the
middle
of theofcurve.
heights
18 yr. The
old men
shape of the curve is
standardized
scores
determined
by thetest
standard
deviation  .
lengths of pregnancies
time for corn to pop
Normal Distribution:
13.5%
34%
3 2  
2.35%
The area under the curve
from a to b represents the
probability of an event
occurring within that range.
 2 3
“68, 95, 99.7 rule”
If we know the equation of the curve we can use calculus to
determine probabilities:
Normal Probability
Density Function:
(Gaussian curve)
2
2

x


/
2



1


f  x 
e
 2
Normal Distribution:
The good news is that you do not have to memorize this
equation!
In real life, statisticians rarely see this function. They use
computer programs or graphing calculators with statistics
software to draw the curve or predict the probabilities.
Normal Probability
Density Function:
(Gaussian curve)
2
2

x


/
2



1


f  x 
e
 2