9 work x

Work
The work W that is done by a constant force F
moving an object a distance d in the direction of the
force is defined to be W = F * d

Work
For example, a book that weights 10 lb is moved
vertically a distance of 5 ft.

Work
vertically a distance of 5 ft. The work that's done is
W = 5*10 = 50 (ft-lb)

Work
We want to calculate the work done on an object
under a nonconstant force.
W = 5*10 = 50 (ft-lb)

Work
under a nonconstant force. Some examples are:
* tension of a stretched spring
* gravitational force
* magnetic force
W = 5*10 = 50 (ft-lb)

Work
under a nonconstant force. Some examples are:
* tension of a stretched spring
* gravitational force
* magnetic force
The tension of a stretched (or compressed) spring
is given by the Hooke's Law.
W = 5*10 = 50 (ft-lb)

Work
Hook's Law:
A spring is stretched by a distance x
from a relaxed position, the tension-
force F on the spring is
F = kx
where k is called the spring constant.
relaxed stretched
x

Work
Hook's Law:
F = kx
relaxed stretched
x
The spring constant is based on factors such as the
material or the size of the spring .

Work
Hook's Law:
F = kx
relaxed stretched
x
Example A: The tension on a bungee cord stretched
5 ft from a relaxed position is 15 lb,
a. find the spring constant.

Work
Hook's Law:
F = kx
relaxed stretched
x
Since15 = k*5 so k = 3 lb/ft is the spring constant.

Work
Hook's Law:
F = kx
relaxed stretched
x
Since15 = k*5 so k = 3 lb/ft is the spring constant.
b. find the tension when it is stretched 25 ft.

Work
Hook's Law:
F = kx
relaxed stretched
x
Since15 = k*5 so k = 3 is lb/ft the spring constant.
b. find the tension when it is stretched 25 ft.
F = 3x. If x = 25, F = 3*25 = 75 lb.

Work
We are to find the work done in extending a spring
stretched from x = a to x = b.

Work
a b
0

Work
a b
Let the sequence {a=x0, x1, x2, .. ,xn=b} be an equi-
partition of [a, b].
0

Work
a=x0 b=xn
x i-1 x ix1 x2
x n-1
0

Work
a=x0 b=xn
x i-1 x ix1 x2
x n-1
Let Δx be the length of each subinterval.
0
Δx

Work
The tension force is kxi-1 when its stretched to xi-1 .
a=x0 b=xn
x i-1 x ix1 x2
x n-10
Δx

Work
a=x0 b=xn
x i-1 x ix1 x2
x n-1
Δx
0

Work
a=x0 b=xn
x i-1 x ix1 x2
x n-1
Extending it to xi,
Δx
0

Work
a=x0 b=xn
x i-1 x ix1 x2
x n-1
Extending it to xi, the work required is approximately
equal to (force)*(distance) = kxi-1* Δx.
Δx
0

Work
Hence the total amount of work required to extend
the spring from x = a to b is approximately the sum
kx0Δx + kx1Δx+ …* + kxn-1Δx =
n
∑ kxiΔx
i=0

Work
The exact work W needed is the limit of these sums
as Δx0,
kx0Δx + kx1Δx+ …* + kxn-1Δx = ∑ kxiΔx
n
i=1

Work
∑ kxi Δx
as Δx0, i.e.
W = lim
n
i=1
n∞

Work
∑ kxi Δx
as Δx0, i.e.
W = limn∞
By FTC, this is
the definite integral
from a to b,
n
i=1

Work
∑ kxi Δx
as Δx0, i.e.
W = lim By FTC, this is
from a to b,
= ∫x=a
b
n
i=1
n∞

Work
∑ kxi Δx
as Δx0, i.e.
from a to b,
= ∫x=a
b
kx dx
n
i=1
n∞

Work
∑ kxi Δx
as Δx0, i.e.
from a to b,
= ∫x=a
b
kx dx
Theorem (Spring Work): The work W needed to
extend a spring from x = a to b is W = ∫x=a
b
kx dx.
n
i=1
n∞

Work
Example B. From a relaxed position, the tension on
a cord extended 5 ft is 15 lb. Find the work needed
to extend it from 4 ft to 20 ft.

Work
Since15 = k*5 so k = 3 is the spring constant.

Work
Hence the force function is F = 3x.

Work
The work required to extend it from 4 ft to 20 ft is

Work
W = ∫x=4
20
3x dx

Work
W = ∫x=4
20
3x dx
=
3x2
2 |
x=4
20

Work
W = ∫x=4
20
3x dx
=
3x2
2 |
x=4
20
=
3
2
(202 – 42) = 576 (ft-lb)

Work
Next, we want to calculate the
work required in pumping a
body of water to a height h
above the container as shown.

Work
h
H2O

Work
h
We will fix all the distance
measurements in feet and
let ρ (= 62.4 lb/ft3) represents
the density of water.
H2O

Work
h
Set the x-axis as shown with x = 0 at the top of the
volume with x = b at the bottom of the volume so x
measures the depth.
x = 0
x = b
H2O

Work
h
Set the x-axis as shown with x = 0 at the top of the
volume with x = b at the bottom of the volume so x
measures the depth.
Let the A(x) be the area of the cross section at depth
x as shown.
x = 0
x = b
xA(x)
H2O

Work
h
0
b
Let the sequence
{0=x0, x1, x2, .. ,xn=b} be an equi-
partition of [0, b]
H2O

Work
h
x0 = 0
xn = b
xi
Let the sequence
{0=x0, x1, x2, .. ,xn=b} be an equi-
partition of [0, b]
x1
x2H2O

Work
h
x0 = 0
xn = b
xiA(xi)
Let the sequence
{0=x0, x1, x2, .. ,xn=b} be an equi-
partition of [0, b] and let the cross-
sectional area at xi be A(xi).
x1
x2H2O

Work
h
x0 = 0
xn = b
xiA(xi)
Let the sequence
{0=x0, x1, x2, .. ,xn=b} be an equi-
Let Δx be the length of each
subinterval.
x1
x2H2O

Work
h
x0 = 0
xn = b
xiA(xi)
Let the sequence
{0=x0, x1, x2, .. ,xn=b} be an equi-
subinterval.
x1
x2
h
x0 = 0
xn = b
xiA(xi)
xi-1
Δx
From depth xi-1 to xi, there is a
"slice" of water whose volume
is approximately A(xi)Δx.
H2O

Work
h
x0 = 0
xn = b
xiA(xi)
Let the sequence
{0=x0, x1, x2, .. ,xn=b} be an equi-
subinterval.
x1
x2
h
x0 = 0
xn = b
xiA(xi)
xi-1
Δx
So the weight of this slice is
approximately ρ*A(xi)Δx.
H2O

Work
h
x0 = 0
xn = b
xiA(xi)
Let the sequence
{0=x0, x1, x2, .. ,xn=b} be an equi-
subinterval.
x1
x2
h
x0 = 0
xn = b
xiA(xi)
xi-1
Δx
So the weight of this slice is
approximately ρ*A(xi)Δx.
Weight of this slice  ρ*A(xi)Δx
H2O

Work
We need to lift this slice of
water (h + xi) feet vertically.
h
x0 = 0
xn = b
xiA(xi)
xi-1
Δx
weight  ρ*A(xi)Δx
h + xi

Work
Therefore the work required
to lift this slice is
approximately equal to
(weight)*(distance)
= (ρA(xi)Δx)*(h + xi)
h
x0 = 0
xn = b
xiA(xi)
xi-1
Δx
h + xi

Work
So the work required to lift all the slices is roughly:
(weight)*(distance)
ρ*A(x1)(h+x1)Δx+ρ*A(x2)(h+x2)Δx+..+ ρ*A(xn)(h+xn)Δx
h
x0 = 0
xn = b
xiA(xi)
xi-1
Δx
h + xi

Work
(weight)*(distance)
= Σ ρ*A(xi)(h+xi)Δx
i=1
n
h
x0 = 0
xn = b
xiA(xi)
xi-1
Δx
h + xi

Work
(weight)*(distance)
i=1
n
h
x0 = 0
xn = b
xiA(xi)
xi-1
Δx
h + xi
A(xi) = cross-sectional area

Work
(weight)*(distance)
i=1
n
h
x0 = 0
xn = b
xiA(xi)
xi-1
Δx
h + xi
A(xi)Δx  volume of the slice

Work
(weight)*(distance)
i=1
n
h
x0 = 0
xn = b
xiA(xi)
xi-1
Δx
h + xi
ρ*A(xi)Δx  weight of the slice

Work
(weight)*(distance)
i=1
n
h
x0 = 0
xn = b
xiA(xi)
xi-1
Δx
h + xi
ρ*A(xi)(h+xi)Δx  work needed to lift the slice
ρ*A(xi)Δx  weight of the slice

Work
The exact work W is the limit of this sum as Δx  0,
W = lim Σ ρ*A(xi)(h+xi)Δx
i=1
n
n∞
h
x = 0
x = b
A(x)

Work
i=1
n
By FTC, this is
from 0 to b,
n∞
h
x = 0
x = b
A(x)

Work
i=1
n
By FTC, this is
from 0 to b,
= ∫x=0
b
n∞
h
x = 0
x = b
A(x)

Work
i=1
n
By FTC, this is
from 0 to b,
= ∫x=0
b
ρA(x)(h+x) dx
n∞
h
x = 0
x = b
A(x)

Work
i=1
n
By FTC, this is
from 0 to b,
= ∫x=0
b
ρA(x)(h+x) dx
= ∫x=0
b
A(x)(h+x) dxρ
n∞
h
x = 0
x = b
A(x)

Work
i=1
n
By FTC, this is
from 0 to b,
= ∫x=0
b
ρA(x)(h+x) dx
h
x = 0
x = b
A(x)
= ∫x=0
b
A(x)(h+x) dxρ
n∞
x-sectional area
x

Work
i=1
n
By FTC, this is
from 0 to b,
= ∫x=0
b
ρA(x)(h+x) dx
h
x = 0
x = b
= ∫x=0
b
A(x)(h+x) dxρ
n∞
x-sectional area distance it moved
A(x)
x
h + x

Work
i=1
n
By FTC, this is
from 0 to b,
= ∫x=0
b
ρA(x)(h+x) dx
h
x = 0
x = b
= ∫x=0
b
A(x)(h+x) dxρ
n∞
x-sectional area
Following are examples of
solids and their cross-
sectional area functions A(x).
A(x)
h + x
x
distance it moved

Work
Example C. Find A(x) of the given shapes.

Work
Example B. Find A(x) of the given shapes.
a. Cylinder of radius R.
R
A(x)

Work
R
A(x)The cross-sections are circles.
So A(x) = πR2
0
x

Work
R
So A(x) = πR2
b. Semi–sphere of radius R.
0
x

Work
R
So A(x) = πR2
Cross-sections are circles
R
A(x)
0
x

Work
R
So A(x) = πR2
0
x
R
R
r
0
x
x
r

Work
R
So A(x) = πR2
0
x
with radius R2 – x2.
R
R
r =R2 – x2
r
0
x
x

Work
R
So A(x) = πR2
0
x
with radius R2 – x2.
Hence A(X) = π(R2 – x2) R
R
r
0
x
x
r =R2 – x2

Work
Example D. Find the work required to pump the
water out of a filled semi sphere container with 10-ft
radius over the rim.

Work
10

Work
10
A(x)

Work
10
A(x)
0
10

Work
r
x
0
10
at depth x

Work
0
x
10
x 10
0
10
r
at depth x
r=100 – x2

Work
0
x
Hence A(X) = π(100 –x2)
10
r=100 – x2
x 10
A(X) = π(100 –x2)
with radius 100 – x2. 0
10
at depth x

Work
0
x
Hence A(X) = π(100 –x2)
10
Volume of a thin slice is
π(100 –x2)Δx and it weights ρπ(100 – x2)Δx.
x 10
A(X) = π(100 –x2)
10
at depth x
r=100 – x2

Work
0
x
Hence A(X) = π(100 –x2)
10
The work to move this slice is ρπ(100 – x2)Δx*x
x 10
A(X) = π(100 –x2)
10
at depth x
r=100 – x2

Work
0
x
Hence A(X) = π(100 –x2)
10
The depth of water is from x = 0 to x = 10
x 10
A(X) = π(100 –x2)
10
at depth x
r=100 – x2

Work
So the work needed is W =
0
x
Hence A(X) = π(100 –x2)
10
∫x=0
10
ρ π(100 –x2)*xdx
x 10
A(X) = π(100 –x2)
10
at depth x
r=100 – x2

Work
So the work needed is W =
0
x
Hence A(X) = π(100 –x2)
10
∫x=0
10
ρ π(100 –x2)*xdx
= 2500πρ
x 10
A(X) = π(100 –x2)
10
r=100 – x2
at depth x

Work
The placement of the measurement x is not fixed,
algebra may be simplified with different placement
of the x-axis.

Work
of the x-axis.
Example E. Find the work required to pumped all the
water in the container as shown to 10 ft above it's rim.

Work
Panels changed: 3,7,12,15,43,58,74,87.
In panels 16 - 23, x should = 0 at the relaxed
right end of the spring. So a = xo = 0.
In panel 24 the upper limit of the sum should
be n - 1.

Work
of the x-axis.
6
A(x)
8
Example E. Find the work required to pump all the
10

Work
of the x-axis.
6
8
0
x
6
A(x) r8
Place the x-measurement from bottom to top.
10

Work
of the x-axis.
6
A(x)
8
10
A(x) is a circle with
radius r where
r
x
= 6
8
8
0
x
6
r

Work
of the x-axis.
6
A(x)
8
10
A(x) is a circle with
radius r where
r
x
= 6
8
r = 3x
4
8
0
x
6
r

Work
6
A(x)
8
10
So A(x) = π(3x/4) 2
=9πx2/16
8
0
x
6
r = 3x/4

Work
6
A(x)
8
10
Volume of a slice
 9πx2/16Δx 8
0
x
6
So A(x) = π(3x/4) 2
=9πx2/16
r = 3x/4

Work
6
A(x)
8
10
Volume of a slice
 9πx2/16Δx
The slice is to be lifted by
a distance of (18 – x).
8
0
x
6
So A(x) = π(3x/4) 2
=9πx2/16
r = 3x/4
(18–x)

Work
6
A(x)
8
10
Volume of a slice
 9πx2/16Δx
Hence the work W needed is
∫x=0
8
9πx2
(18 – x)dxW = 16
8
0
x
6
So A(x) = π(3x/4) 2
=9πx2/16
r = 3x/4
ρ
(18–x)

Work
6
A(x)
8
10
Volume of a slice
 9πx2/16Δx
∫x=0
8
9πx2
(18 – x)dxW = 16
8
0
x
6
So A(x) = π(3x/4) 2
=9πx2/16
r = 3x/4
= ∫x=0
8
9πρ x2*(18 – x)dx
16
ρ
(18–x)

Work
6
A(x)
8
10
Volume of a slice
 9πx2/16Δx
∫x=0
8
9πx2
(18 – x)dxW = 16
8
0
x
6
So A(x) = π(3x/4) 2
=9πx2/16
r = 3x/4
= ∫x=0
8
9πρ x2*(18 – x)dx =
16
ρ
(18–x)
9πρ
16 (6x3 – x4/4)|
x=0
8

Work
6
A(x)
8
10
Volume of a slice
 9πx2/16Δx
∫x=0
8
9πx2
(18 – x)dxW = 16
8
0
x
6
So A(x) = π(3x/4) 2
=9πx2/16
r = 3x/4
(18–x)
= ∫x=0
8
9πρ x2*(18 – x)dx =
16
ρ
9πρ
16 (6x3 – x4/4)|
x=0
8
x3(6 – x/4)

Work
6
A(x)
8
10
Volume of a slice
 9πx2/16Δx
∫x=0
8
9πx2
(18 – x)dxW = 16
8
0
x
6
So A(x) = π(3x/4) 2
=9πx2/16
r = 3x/4
(18–x)
= 1152πρ
= ∫x=0
8
9πρ x2*(18 – x)dx =
16
ρ
9πρ
16 (6x3 – x4/4)|
x=0
8
x3(6 – x/4)

9 work x

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