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Conservation of Energy with Dissipative Forces
Introduction
We earlier saw W NC = /\KE
+ /\PE We can use this to calculate
information about the nonconservative forces. Remember Wfr
= - Ffrd (It is negative because the direction of
the force is the opposite of the displacement) Table of Contents Problems
| 1. |
A 1.00 x 10 4 kg truck is at the
top of a 1000. m mountain. When is allowed to use coast down this
mountain and up the next mountain, it only reaches a height of 850.
m. The truck travels 3000. m, what is the average force of friction?
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| 2. |
A basketball, 2.00 kg, is dropped from the
top of the Hover Dam, height of 120. m, it is only falling at
29.9 m/s when it hits. What is the average force of friction?
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| 3. |
A man is on a sled at the top of a hill that
is 50.0 m high and 120. m long. At the bottom of the hill, the man
and sled are moving 20.0 m/s. If the total mass of the man and sled
is 110. kg, what is the coefficient of kinetic friction?
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| 4. |
The average force of friction on a softball,
1.00 kg, dropping from the top of the Empire State Building is 2.00 N. If the
softball drops 300. m, how fast is it falling? |
Table of Contents Answers
| 1. |
A 1.00 x 10 4 kg truck is at the
top of a 1000. m mountain. When is allowed to use coast down this
mountain and up the next mountain, it only reaches a height of 850.
m. The truck travels 3000. m, what is the average force of friction?
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|
Since there is no Kinetic Energy at top of either mountain, WNC
= /\PE.
WNC = Wfr = - Ffrd
= mgh1 - mgh2 = mg (h1
- h2)
mg (h1 - h2)
1.00 x 10 4 kg x 9.80 m s -2 ( 1000. m - 850. m )
Ffr = ---------------------- =
--------------------------------------------------------------- = 4.90 x
10 3 N
-
d
3000. m
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|
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| 2. |
A basketball, 2.00 kg, is dropped from the
top of the Hover Dam, height of 120. m, it is only falling at
29.9 m/s when it hits. What is the average force of friction?
|
|
There is only Potential Energy at the top.
Only Kinetic Energy is present at the bottom.
WNC = Wfr = - Ffrd
= /\PE + /\KE = mgh1 -  mv22
mg h1 - mv22
2.00 kg x 9.80 m s -2 x 120. m -
x 2.00 kg x ( 29.9 m s
-1)2
Ffr = ---------------------- =
-------------------------------------------------------------------------------
-
d
120. m
Ffr = 12.5 N
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|
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| 3. |
A man is on a sled at the top of a hill that
is 50.0 m high and 120. m long. At the bottom of the hill, the man
and sled are moving 20.0 m/s. If the total mass of the man and sled
is 110. kg, what is the coefficient of kinetic friction?
|
|
There is only Potential Energy at the top.
Only
Kinetic Energy is present at the bottom.
WNC = Wfr = - Ffrd
= /\PE + /\KE = mgh1
- mv22
mg h1 - mv22
110. kg x 9.80 m s -2 x 50.0 m -
x 110. kg x ( 20.0 m s
-1 ) 2
Ffr = ----------------------- = --------------------------------------------------------------------------------
-
d
120. m
Ffr = 264 N
50.0 m
 = Sin
-1 ( ---------- ) = 24.6o
120. m
Ffr =  FN
= FwCos
Ffr
264 N
= ------------ = ------------------------------------------
FwCos
110. kg x 9.80 m s -2 Cos 24.6o
= 0.269
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| 4. |
The average force of friction on a softball,
1.00 kg, dropping from the top of the Empire State Building is
2.00 N. If the
softball drops 300. m, how fast is it falling?
|
|
There is only Potential Energy at the top.
Only
Kinetic Energy is present at the bottom.
WNC = Wfr = - Ffrd
= /\PE + /\KE = mgh1
- mv22
2mgh1 + 2Ffrd
2 x 1.00 kg x 9.80 m s -2 x 300. m + 2 x 2.00 N ( - 300 m)
v = ( --------------------- ) 1/2 = ( --------------------------------------------------------------------------- )
1/2 = 68.4 m s -1
m
1.00 kg
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Table of Contents |
