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Work=
------=_NextPart_01C97CE4.362127A0 Content-Location: file:///C:/CD8D4AAB/6Work_files/slide0034.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Slide 1
Work=
The work W that is done by a constant force F moving an object of a distance d is defined to be=
        = ;            &n= bsp;           W =3D F * d 
------=_NextPart_01C97CE4.362127A0 Content-Location: file:///C:/CD8D4AAB/6Work_files/slide0035.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Slide 1
Work=
The work W that is done by a constant force F moving an object of a distance d is defined to be=
        = ;            &n= bsp;           W =3D F * d 
Example:  A bo= ok that weights 10 lb is moved vertically by a distance of 5 ft.
------=_NextPart_01C97CE4.362127A0 Content-Location: file:///C:/CD8D4AAB/6Work_files/slide0036.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Slide 1
Work=
The work W that is done by a constant force F moving an object of a distance d is defined to be=
        = ;            &n= bsp;           W =3D F * d 
Example:  A bo= ok that weights 10 lb is moved vertically by a distance of 5 ft. The work that done is        = ;          
        = ;            &n= bsp;       W =3D 5*= 10 =3D 50 (ft-lb) 
------=_NextPart_01C97CE4.362127A0 Content-Location: file:///C:/CD8D4AAB/6Work_files/slide0037.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Slide 1
Work=
The work W that is done by a constant force F moving an object of a distance d is defined to be=
        = ;            &n= bsp;           W =3D F * d 
Example:  A bo= ok that weights 10 lb is moved vertically by a distance of 5 ft. The work that done is        = ;          
        = ;            &n= bsp;       W =3D 5*= 10 =3D 50 (ft-lb) 
We want to calculate the work done on an object = under a = nonconstant force.
------=_NextPart_01C97CE4.362127A0 Content-Location: file:///C:/CD8D4AAB/6Work_files/slide0101.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Slide 1
Work=
The work W that is done by a constant force F moving an object of a distance d is defined to be=
        = ;            &n= bsp;           W =3D F * d 
Example:  A bo= ok that weights 10 lb is moved vertically by a distance of 5 ft. The work that done is        = ;          
        = ;            &n= bsp;       W =3D 5*= 10 =3D 50 (ft-lb) 
We want to calculate the work done on an object = under a = nonconstant force. Some examples are:=
* tension of a stretched spring
* gravitational force
* magnetic force
------=_NextPart_01C97CE4.362127A0 Content-Location: file:///C:/CD8D4AAB/6Work_files/slide0102.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Slide 1
Work=
The work W that is done by a constant force F moving an object of a distance d is defined to be=
        = ;            &n= bsp;           W =3D F * d 
Example:  A bo= ok that weights 10 lb is moved vertically by a distance of 5 ft. The work that done is        = ;          
        = ;            &n= bsp;       W =3D 5*= 10 =3D 50 (ft-lb) 
We want to calculate the work done on an object = 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 Hook's Law.
------=_NextPart_01C97CE4.362127A0 Content-Location: file:///C:/CD8D4AAB/6Work_files/slide0040.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Slide 1
Work=
relaxed
stretched
x
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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 =3D kx
where k is called the spring constant.
relaxed
stretched
x
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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 =3D kx
where k is called the spring constant.
relaxed
stretched
x
The spring constant is based on factors such as the material, the size of the spring . =
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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 =3D kx
where k is called the spring constant.
relaxed
stretched
x
The spring constant is based on factors such as the material, the size of the spring . =
Example: The tension on a bungee co= rd stretched
5 ft from a relaxed position is 15 = lb,
a. find the spring constant.
&n= bsp;
        = ;  
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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 =3D kx
where k is called the spring constant.
relaxed
stretched
x
The spring constant is based on factors such as the material, the size of the spring . =
Example: The tension on a bungee co= rd stretched
5 ft from a relaxed position is 15 = lb,
a. find the spring constant.
Since15 =3D k*5 so k =3D 3 is the s= pring constant.
   
        = ;  
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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 =3D kx
where k is called the spring constant.
relaxed
stretched
x
The spring constant is based on factors such as the material, the size of the spring . =
Example: The tension on a bungee co= rd stretched
5 ft from a relaxed position is 15 = lb,
a. find the spring constant.
Since15 =3D k*5 so k =3D 3 is the s= pring constant.
b. find the tension when it is stre= tched 25 ft.
        = ;  
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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 =3D kx
where k is called the spring constant.
relaxed
stretched
x
The spring constant is based on factors such as the material or the size of the spring .
Example: The tension on a bungee co= rd stretched
5 ft from a relaxed position is 15 = lb,
a. find the spring constant.
Since15 =3D k*5 so k =3D 3 is the s= pring constant.
b. find the tension when it is stre= tched 25 ft.
F =3D 3x. If x =3D 25, F =3D 3*25 = =3D 75 lb.   
        = ;  
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Work=
We are to find the work done in extending a spring stretched by x =3D a to x =3D b. 
------=_NextPart_01C97CE4.362127A0 Content-Location: file:///C:/CD8D4AAB/6Work_files/slide0050.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Slide 1
Work=
a
b
We are to find the work done in extending a spring stretched by x =3D a to x =3D b. 
0
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Work=
a
b
Let the sequence {a=3Dx0, x<= /span>1= , x2, ..= ,xn= =3Db} be an equi- partition of [a, b]. 
We are to find the work done in extending a spring stretched by x =3D a to x =3D b. 
0
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Work=
a=3Dx0
b=3Dxn
x i-1
x i
x1
x2
x n-1
Let the sequence {a=3Dx0, x<= /span>1= , x2, ..= ,xn= =3Db} be an equi- partition of [a, b]. 
We are to find the work done in extending a spring stretched by x =3D a to x =3D b. 
0
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Work=
a=3Dx0
b=3Dxn
x i-1
x i
x1
x2
x n-1
Let the sequence {a=3Dx0, x<= /span>1= , x2, ..= ,xn= =3Db} be an equi- partition of [a, b]. 
Let = 16;x be the length of each subinterval.
We are to find the work done in extending a spring stretched by x =3D a to x =3D b. 
0
Δx
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Work=
Let the sequence {a=3Dx0, x<= /span>1= , x2, ..= ,xn= =3Db} be an equi- partition of [a, b]. 
Let = 16;x be the length of each subinterval.
The tension force is kxi-1 when its stretched to xi-1<= span style=3D'mso-spacerun:yes'> .
We are to find the work done in extending a spring stretched by x =3D a to x =3D b. 
a=3Dx0
b=3Dxn
x i-1
x i
x1
x2
x n-1
0
Δx
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Work=
a=3Dx0
b=3Dxn
x i-1
x i
x1
x2
x n-1
Let the sequence {a=3Dx0, x<= /span>1= , x2, ..= ,xn= =3Db} be an equi- partition of [a, b]. 
Let = 16;x be the length of each subinterval.
The tension force is kxi-1 when its stretched to xi-1<= span style=3D'mso-spacerun:yes'> .
Δx
We are to find the work done in extending a spring stretched by x =3D a to x =3D b. 
0
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Work=
a=3Dx0
b=3Dxn
x i-1
x i
x1
x2
x n-1
Extend= ing it to xi, <= /div>
Let the sequence {a=3Dx0, x<= /span>1= , x2, ..= ,xn= =3Db} be an equi- partition of [a, b]. 
Let = 16;x be the length of each subinterval.
The tension force is kxi-1 when its stretched to xi-1<= span style=3D'mso-spacerun:yes'> .
Δx
We are to find the work done in extending a spring stretched by x =3D a to x =3D b. 
0
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Work=
a=3Dx0
b=3Dxn
x i-1
x i
x1
x2
x n-1
Extending it to xi, the work required is approximately <= /div>
equal to (force)*(distance) =3D kxi= -1* = 16;x. <= /div>
Let the sequence {a=3Dx0, x<= /span>1= , x2, ..= ,xn= =3Db} be an equi- partition of [a, b]. 
Let = 16;x be the length of each subinterval.
The tension force is kxi-1 when its stretched to xi-1<= span style=3D'mso-spacerun:yes'> .
Δx
We are to find the work done in extending a spring stretched by x =3D a to x =3D b. 
0
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Work=
Hence= the total amount of work required to extend the spring from x =3D a to b is approximately the= sum
kx0= Δx +  
kx1= Δx+ …  
*
+ kx<= /span>n= -1Δx =3D 
n
ͨ= 1; kxi= Δx  
i=3D= 1
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Work=
Hence the total amount of work required to extend the spring from x =3D a to b is approximately the= sum
The exact work W needed is the limit of these sums = as = 16;xà0,
kx0= Δx +  
kx1= Δx+ …  
*
+ kx<= /span>n= -1Δx =3D 
ͨ= 1; kxi= Δx  
n
i=3D= 1
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Work=
Hence the total amount of work required to extend the spring from x =3D a to b is approximately the= sum
ͨ= 1; kxi= Δx  
The exact work W needed is the limit of these sums = as = 16;xà0, i.e. 
W =3D= lim
kx0= Δx +  
kx1= Δx+ …  
*
+ kx<= /span>n= -1Δx =3D 
ͨ= 1; kxi= Δx  
n
i=3D= 1
n&agr= ave;∞
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Work=
Hence the total amount of work required to extend the spring from x =3D a to b is approximately the= sum
ͨ= 1; kxi= Δx  
The exact work W needed is the limit of these sums = as = 16;xà0, i.e. 
W =3D= lim
n&agr= ave;∞
By FTC, this is
the definite integral
from a to b,
kx0= Δx +  
kx1= Δx+ …  
*
+ kx<= /span>n= -1Δx =3D 
ͨ= 1; kxi= Δx  
n
i=3D= 1
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Work=
Hence the total amount of work required to extend the spring from x =3D a to b is approximately the= sum
ͨ= 1; kxi= Δx  
The exact work W needed is the limit of these sums = as = 16;xà0, i.e. 
W =3D= lim
By FTC, this is
the definite integral
from a to b,
=3D
W= 47;
x=3D= a
b
kx0= Δx +  
kx1= Δx+ …  
*
+ kx<= /span>n= -1Δx =3D 
ͨ= 1; kxi= Δx  
n
i=3D= 1
n&agr= ave;∞
------=_NextPart_01C97CE4.362127A0 Content-Location: file:///C:/CD8D4AAB/6Work_files/slide0059_image136.gif Content-Transfer-Encoding: base64 Content-Type: image/gif R0lGODlhDAAaAHcAMSH+GlNvZnR3YXJlOiBNaWNyb3NvZnQgT2ZmaWNlACH5BAEAAAAALAMAAgAG ABUAgAAAAAAAAAIYBBJmqJfK4FvTtUhvxVbq7yDimIwUxkUFADs= ------=_NextPart_01C97CE4.362127A0 Content-Location: file:///C:/CD8D4AAB/6Work_files/slide0060.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Slide 1
Work=
Hence the total amount of work required to extend the spring from x =3D a to b is approximately the= sum
ͨ= 1; kxi= Δx  
The exact work W needed is the limit of these sums = as = 16;xà0, i.e. 
W =3D= lim
By FTC, this is
the definite integral
from a to b,
=3D
W= 47;
x=3D= a
b
 kx dx  
kx0= Δx +  
kx1= Δx+ …  
*
+ kx<= /span>n= -1Δx =3D 
ͨ= 1; kxi= Δx  
n
i=3D= 1
n&agr= ave;∞
------=_NextPart_01C97CE4.362127A0 Content-Location: file:///C:/CD8D4AAB/6Work_files/slide0060_image137.gif Content-Transfer-Encoding: base64 Content-Type: image/gif R0lGODlhQAAaAHcAMSH+GlNvZnR3YXJlOiBNaWNyb3NvZnQgT2ZmaWNlACH5BAEAAAAALAEAAgA8 ABYAgAAAAAAAAAJrBBKmy83ookRJQYlfzZmuy3VbGHnWSDpg2pjGymqxusLzi97uket9vLP5bpqR cHZkBX9AZmpJlEVxRufTGoJOcVtetcvrarfJa4Wya6KJ6Paa/Yab4nIspk2138uk8ZSfVaPXAXZQ eNiiUAAAOw== ------=_NextPart_01C97CE4.362127A0 Content-Location: file:///C:/CD8D4AAB/6Work_files/slide0061.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Slide 1
Work=
Hence the total amount of work required to extend the spring from x =3D a to b is approximately the= sum
ͨ= 1; kxi= Δx  
The exact work W needed is the limit of these sums = as = 16;xà0, i.e. 
W =3D= lim
By FTC, this is
the definite integral
from a to b,
=3D
W= 47;
x=3D= a
b
 kx dx  
Theorem (Spring Work): The work W needed to
  = ;
extend a spring from x =3D a to b is W =3D=
W= 47;
x=3D= a
b
 kx dx.  
kx0= Δx +  
kx1= Δx+ …  
*
+ kx<= /span>n= -1Δx =3D 
ͨ= 1; kxi= Δx  
n
i=3D= 1
n&agr= ave;∞
------=_NextPart_01C97CE4.362127A0 Content-Location: file:///C:/CD8D4AAB/6Work_files/slide0061_image138.gif Content-Transfer-Encoding: base64 Content-Type: image/gif R0lGODlhQAAaAHcAMSH+GlNvZnR3YXJlOiBNaWNyb3NvZnQgT2ZmaWNlACH5BAEAAAAALAEAAgA8 ABYAgAAAAAAAAAJrBBKmy83ookRJQYlfzZmuy3VbGHnWSDpg2pjGymqxusLzi97uket9vLP5bpqR cHZkBX9AZmpJlEVxRufTGoJOcVtetcvrarfJa4Wya6KJ6Paa/Yab4nIspk2138uk8ZSfVaPXAXZQ eNiiUAAAOw== ------=_NextPart_01C97CE4.362127A0 Content-Location: file:///C:/CD8D4AAB/6Work_files/slide0018.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Slide 1
Work=
Example: From a relaxed position, the tension on a <= /span>cord extended 5 ft is 15 lb. Find t= he work needed to extend it from 4 ft to 20 ft.
------=_NextPart_01C97CE4.362127A0 Content-Location: file:///C:/CD8D4AAB/6Work_files/slide0062.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Slide 1
Work=
Example: From a relaxed position, the tension on a <= /span>cord extended 5 ft is 15 lb. Find t= he work needed to extend it from 4 ft to 20 ft.
Since15 =3D k*5 so k =3D 3 is the s= pring constant.
------=_NextPart_01C97CE4.362127A0 Content-Location: file:///C:/CD8D4AAB/6Work_files/slide0063.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Slide 1
Work=
Example: From a relaxed position, the tension on a <= /span>cord extended 5 ft is 15 lb. Find t= he work needed to extend it from 4 ft to 20 ft.
Since15 =3D k*5 so k =3D 3 is the s= pring constant.
Hence the force function is F =3D 3= x.
------=_NextPart_01C97CE4.362127A0 Content-Location: file:///C:/CD8D4AAB/6Work_files/slide0064.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Slide 1
Work=
Example: From a relaxed position, the tension on a <= /span>cord extended 5 ft is 15 lb. Find t= he work needed to extend it from 4 ft to 20 ft.
Since15 =3D k*5 so k =3D 3 is the s= pring constant.
Hence the force function is F =3D 3= x.
The work required to extend it from= 4 ft to 20 ft is
------=_NextPart_01C97CE4.362127A0 Content-Location: file:///C:/CD8D4AAB/6Work_files/slide0065.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Slide 1
Work=
Example: From a relaxed position, the tension on a <= /span>cord extended 5 ft is 15 lb. Find t= he work needed to extend it from 4 ft to 20 ft.
Since15 =3D k*5 so k =3D 3 is the s= pring constant.
Hence the force function is F =3D 3= x.
The work required to extend it from= 4 ft to 20 ft is
W =3D
∫
x=3D4
20
 3x dx  
------=_NextPart_01C97CE4.362127A0 Content-Location: file:///C:/CD8D4AAB/6Work_files/slide0066.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Slide 1
Work=
Example: From a relaxed position, the tension on a <= /span>cord extended 5 ft is 15 lb. Find t= he work needed to extend it from 4 ft to 20 ft.
Since15 =3D k*5 so k =3D 3 is the s= pring constant.
Hence the force function is F =3D 3= x.
The work required to extend it from= 4 ft to 20 ft is
W =3D
∫
x=3D4
20