“WORK” AS AN APPLICATION OF THE DEFINITE INTEGRAL
Typing this document requires a lot of “work”. If said in the presence of a physicist, they might
wonder as to what planet does this moron come from?
WORK DONE BY A CONSTANT FORCE ACTING ALONG A STRAIGHT LINE
In the simplest case a body undergoes a displacement (i.e., a change in its position) with
magnitude 𝑠𝑠 along a straight line while a constant net force with magnitude 𝐹𝐹, directed alone
the same line, acts on it. We define the work done on the body by a net force of magnitude 𝐹𝐹
as
𝑾𝑾 = Work done by𝑭𝑭 = 𝑭𝑭 ⋅ 𝒔𝒔
The constant net Force
(1)
Direction of motion
m
m
A
B
Line of action of the force
Magnitude of displacement = Distance between A and B
The metric units of work is Newton-meters (Nm) or Joules (J). In British units, work is usually
measured in foot-pounds (ft-lbs)
Conversion:
1 Joule = 1 𝑁𝑁𝑁𝑁 ≈ 0.7376 ft⋅lbs
1 ft⋅lb ≈ 1.356 Nm = 1.356 J
Problem 1: How much work is done when a constant force of 500 𝑁𝑁 is used to
push the blue rectangular cart above at an initial position A through a
displacement of magnitude 25 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 to final position 𝐵𝐵 along the 𝑥𝑥-axis in the
figure above.
Solution:
Amount of Work is 𝑾𝑾 = 𝑭𝑭 ⋅ 𝒅𝒅 = (𝟓𝟓𝟓𝟓𝟓𝟓 𝑵𝑵)(𝟐𝟐𝟐𝟐 𝒎𝒎) = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝑵𝑵𝑵𝑵 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝑱𝑱𝑱𝑱𝑱𝑱𝑱𝑱𝑱𝑱𝑱𝑱
WORK DONE BY A VARYING FORCE ACTING ALONG A STRAIGHT LINE ( Our Main
Concern!)
In this discussion, we consider the work done by a variable force, which is a function of the
position of the object on which the force is acting. We wish to define what is meant by the term
"work" in such a case.
Suppose that, where 𝐹𝐹 is continuous on the interval [𝑎𝑎, 𝑏𝑏], is the magnitude of the force
acting in the direction of motion on an object as it moves to the right along the 𝑥𝑥-axis from
point 𝑥𝑥 = 𝑥𝑥1 to point𝑥𝑥 = 𝑥𝑥2 .
𝑦𝑦 = 𝐹𝐹 (𝑥𝑥)
𝐹𝐹(𝜉𝜉𝑖𝑖 )
𝑥𝑥𝑖𝑖−1
𝜉𝜉𝑖𝑖
𝑥𝑥𝑖𝑖
SLICE, APPROXIMATE, and INTEGRATE PROCEDURE: Let 𝑃𝑃 be a partition of the closed interval
[𝑎𝑎, 𝑏𝑏]:
𝑎𝑎 = 𝑥𝑥0 < 𝑥𝑥1 < 𝑥𝑥2 < ⋯ < 𝑥𝑥𝑖𝑖−1 < 𝑥𝑥𝑖𝑖 < ⋯ < 𝑥𝑥𝑛𝑛−1 < 𝑥𝑥𝑛𝑛 = 𝑏𝑏
The 𝑖𝑖th subinterval is [𝑥𝑥𝑖𝑖−1 , 𝑥𝑥𝑖𝑖 ], and if 𝑥𝑥𝑖𝑖−1 is close to 𝑥𝑥𝑖𝑖 , the force is almost constant in this
subinterval. If we assume the force is constant in the 𝑖𝑖th subinterval and if 𝜉𝜉𝑖𝑖 is any point such
that 𝑥𝑥𝑖𝑖−1 ≤ 𝜉𝜉𝑖𝑖 ≤ 𝑥𝑥𝑖𝑖 , then if Type equation here. is the measure of the work done on the object as
it moves from the point 𝑥𝑥𝑖𝑖−1 to the point 𝑥𝑥𝑖𝑖 , from formula (1) we have Δ𝑊𝑊𝑖𝑖 = 𝐹𝐹 (𝜉𝜉𝑖𝑖 )(𝑥𝑥𝑖𝑖 −
𝑥𝑥𝑖𝑖−1 ). Replacing 𝑥𝑥𝑖𝑖 − 𝑥𝑥𝑖𝑖−1 by Δ𝑥𝑥𝑖𝑖 , we have Δ𝑊𝑊𝑖𝑖 = 𝐹𝐹(𝜉𝜉𝑖𝑖 )Δ𝑥𝑥𝑖𝑖 and
𝑛𝑛
𝑛𝑛
𝑖𝑖=1
𝑖𝑖=1
� Δ𝑊𝑊𝑖𝑖 = � 𝐹𝐹(𝜉𝜉𝑖𝑖 ) Δ𝑥𝑥𝑖𝑖
(2)
The smaller we take the norm (widest subinterval) of the partition 𝑃𝑃, the larger 𝑛𝑛 will be and
the closer the Riemann sum in Eq. (2) will be to what we intuitively think of as the measure of
the total work done. We therefore define the measure of the total work as the limit of the
Riemann sum in Eq. (2).
Definition: Let the function 𝐹𝐹 be continuous on the closed interval [𝑎𝑎, 𝑏𝑏] and
𝐹𝐹(𝑥𝑥) be the magnitude of the force acting on an object at the point 𝑥𝑥 on the 𝑥𝑥axis. Then if 𝑊𝑊 is the work done by the varying force as the object moves from
𝑥𝑥 = 𝑎𝑎 to 𝑥𝑥 = 𝑏𝑏, then 𝑊𝑊 is given by
𝑾𝑾 =
𝐥𝐥𝐥𝐥𝐥𝐥
𝐦𝐦𝐦𝐦𝐦𝐦 𝚫𝚫𝒙𝒙𝒊𝒊 →𝟎𝟎
𝒏𝒏
𝒃𝒃
� 𝑭𝑭(𝝃𝝃𝒊𝒊 )𝚫𝚫𝒙𝒙𝒊𝒊 = � 𝑭𝑭(𝒙𝒙) 𝒅𝒅𝒅𝒅
𝒂𝒂
𝒊𝒊=𝟏𝟏
(𝟑𝟑)
Note however that if 𝐹𝐹 (𝑥𝑥 ) = 𝐹𝐹 (is constant) for all 𝑎𝑎 ≤ 𝑥𝑥 ≤ 𝑏𝑏, then we have
formula (1)
𝑏𝑏
𝑊𝑊 = 𝐹𝐹 ⋅ � 𝑑𝑑𝑑𝑑 = 𝐹𝐹 (𝑏𝑏 − 𝑎𝑎) = 𝐹𝐹 ⋅ 𝑠𝑠
𝑎𝑎
where 𝑠𝑠 = 𝑏𝑏 − 𝑎𝑎 is the length of the interval.
Problem 2: A uniform chain of length 56 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 with a mass of
4 𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘 𝑝𝑝𝑝𝑝𝑝𝑝 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 is dangling from the roof of the NAC-building. Measure
the amount of work that is needed to pull the chain up onto the top of the
building.
Top of the building
x
Solution:
small piece of chain of length Δ𝑥𝑥𝑖𝑖
Nac building
The chain
Since 1 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 of the chain has mass of 4 𝑘𝑘𝑘𝑘, the gravitational force per meter of
chain is (4 𝑘𝑘𝑘𝑘)(9.8𝑚𝑚/𝑠𝑠 2 ) = 39.2 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁. Let us divide the chain into very tiny
pieces of equal length Δ𝑥𝑥, each requiring a force of magnitude 39.2 Δ𝑥𝑥 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁
to move it against gravity. If Δ𝑥𝑥 is very small, all of the pieces is hauled up
approximately the same distance, namely 𝑥𝑥 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 to the top of the building, so
Work done ≈ (39.2 Δ𝑥𝑥 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁)(𝑥𝑥 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 ) = 39.2 𝑥𝑥 Δ𝑥𝑥 𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽𝐽,
Now as Δ𝑥𝑥 → 0, we obtain a definite integral. Since 𝑥𝑥 varies from
0 to 56 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚, the total work is
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑊𝑊ork done
=�
56
0
39.2 2 56
𝑥𝑥 � = (39.2)(1568) = 61465.6 Joules
39.2 𝑥𝑥 𝑑𝑑𝑑𝑑 = �
2
0
Problem 3. A bucket weighing 20 lbs containing 60 lbs of sand is attached to the
lower end of a 100 ft long chain that weighs 10 lbs and is hanging in a deep well.
Find the work done in raising the bucket to the top of the well.
Solution: A good strategy here is to consider the work done in raising only the
bucket and and only the sand to the top of the well; and then finally consider the
work done in moving only the chain to the top of the well.
Raising the bucket alone: To raise the bucket alone to the top of the well, it
would required an amount of work 𝑊𝑊1 given by formula (1) since the force (the
weight of the bucket) is constant. Thus,
𝑊𝑊𝐵𝐵 = (20 lbs)(100 ft)=2000 ft⋅lbs
Raising the sand alone: To raise the sand alone require an amount of work 𝑊𝑊2
given again by formula (1); thus,
𝑊𝑊𝑠𝑠 = (60 lbs)(100 ft)=6000 ft⋅lbs
Raising the chain alone: To raise only the chain we do this similar to in Problem 2
above. The gravitational force per foot of chain is 0.1 pounds. As usual, divide
up the chain into very tiny pieces of equal length Δ𝑥𝑥 feet, each requiring a force of
magnitude 0.1 Δ𝑥𝑥 pounds to move it against gravity. If Δ𝑥𝑥 is very small, all of the
pieces is hauled up to the top of the well approximately the same distance,
namely 𝑥𝑥 feet to the top of the building, so the work Δ𝑊𝑊𝑐𝑐 done on the raising
small piece of the chain to the top of the well is
Δ𝑊𝑊𝑐𝑐 ≈ �
1
𝑥𝑥
Δ𝑥𝑥 pounds� (𝑥𝑥 feet) =
Δ𝑥𝑥 ft-lbs,
10
10
Now as Δ𝑥𝑥 → 0, we obtain a definite integral. Since 𝑥𝑥 varies from 0 to 100 feet,
the work is
100 1
𝑊𝑊𝑐𝑐 = ∫0
10
1
100
𝑥𝑥 𝑑𝑑𝑑𝑑 = � 𝑥𝑥 2 �
20
0
1
= ( )(10000) = 500 ft⋅lbs.
20
Thus the total Work done in raising the bucket to the top of well is
𝑊𝑊 = 𝑊𝑊𝐵𝐵 + 𝑊𝑊𝑠𝑠 + 𝑊𝑊𝑐𝑐 = 2000 ft-lbs +6000 ft⋅lbs + 500 ft-lbs = 8500 ft⋅ lbs
Problem 4 (Lifting a Leaky Bucket). Solve Problem 4, if instead the sand is leaking
out of the bucket at a constant rate and has all leaked out just as soon as the
bucket reaches the top of the well.
Solution: Raise the Bucket alone and then raising the chain alone. This requires a
combined total work (as calculated in Problem 4) of 𝑊𝑊𝐵𝐵 + 𝑊𝑊𝐶𝐶 where
𝑊𝑊𝐵𝐵 + 𝑊𝑊𝐶𝐶 = 2000 ft⋅lbs+500 ft⋅lbs= 2500 ft⋅lbs.
Raising the Sand alone: The force required to raise the sand is equal to the
sand’s weight, which varies steadily from 60 lbs to 0 lbs over the 100 ft lift. When
the bucket is raised 𝑥𝑥 feet, the sand weighs
3
𝐹𝐹 (𝑥𝑥 ) = 60 − 𝑥𝑥 lbs.
5
Therefore, the work done in raising the sand to the top of the well is 𝑊𝑊𝑠𝑠 given by
100
𝑊𝑊𝑠𝑠 = ∫0
3
�60 − 𝑥𝑥� 𝑑𝑑𝑑𝑑 = �60𝑥𝑥 −
5
3
10
100
𝑥𝑥 2 �
0
= 6000 − 3000 = 3000 ft⋅lbs.
Finally, the total work, 𝑊𝑊𝑇𝑇 done in raising the leaky bucket of sand using the chain
is
𝑊𝑊𝑇𝑇 = (𝑊𝑊𝐵𝐵 + 𝑊𝑊𝐶𝐶 ) + 𝑊𝑊𝑆𝑆 = 2500 ft⋅lbs+3000 ft⋅lbs=5500 ft⋅lbs
In case you’re wondering how I obtained the expression for 𝐹𝐹 (𝑥𝑥 )? The sand
leaks out at a steady rate as we were told. This steady rate must be 60 pounds of
sand per 100 feet of chain or equivalently, we leak 3 pounds of sand every 5 feet
3
we raise the chain. Thus, raising chain 𝑥𝑥 feet we leak 𝑥𝑥 pounds of sand and so
5
the sand remaining in the bucket at this point 𝑥𝑥 now weighs
3
𝐹𝐹 (𝑥𝑥 ) = 60 − 𝑥𝑥 pounds (the force).
3
5
Notice: 𝐹𝐹 (𝑥𝑥 ) = 60 − 𝑥𝑥 = 60 ⋅ �1 −
5
1
100
𝑥𝑥� =
60
�
100−𝑥𝑥
⋅ �������
100
original weight
proportion left
of the sand
at elevation𝒙𝒙
Problem 5. A container weighing 4 pounds and a rope of “negligible” weight are
used to lift water to the top of a building which is 80 feet high. The container is
filled with 40 pounds of water and is pulled upward at constant rate of 2 feet per
second, but water leaks from a tiny hold in the container at a constant rate of 1
pound every 5 seconds while it is being pulled upwards. (a) Find the work done in
pulling the filled container to the top of the building. (b) Find the work done if
instead the rope weighs 0.4 pounds per foot.
Solution:
(a) Even though the rope is attached to the container with the water, there is no
work done in pulling the rope alone up to the top of the building.
Work done in pulling the container alone is given by:
𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = Force × Distance = (4 lbs)(80 ft)= 320 ft⋅lbs
The Water “Alone”: Let 𝑥𝑥0 = 0 (measured in feet) denote the position of the 40
pounds water at time 𝑡𝑡0 = 0 (measured in seconds). At time 𝑡𝑡 = 𝑡𝑡0 + Δ𝑡𝑡 = Δ𝑡𝑡
seconds later the water is at position
𝑥𝑥 = 𝑥𝑥0 + Δ𝑥𝑥 = Δ𝑥𝑥; however, during this elapsed time, the weight of water that
leaked out is
1
5
Δ𝑡𝑡 pounds where
Thus, the water 40 −
given by
1
1
10
5
1 1
Δ𝑡𝑡 = � Δ𝑥𝑥� =
5 2
1
10
Δ𝑥𝑥 pounds.
Δ𝑥𝑥 pounds i.e., the force is a function of the position 𝑥𝑥
𝐹𝐹 = 𝐹𝐹 (𝑥𝑥 ) = 40 −
1
𝑥𝑥; 0 ≤ 𝑥𝑥 ≤ 80.
10
Thus, the work done as a function of position is given by
𝑥𝑥 1
𝑊𝑊𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 = � 𝐹𝐹 (𝑥𝑥 ) 𝑑𝑑𝑑𝑑 = �
𝑥𝑥 0
80
0
�40 −
1
𝑥𝑥� 𝑑𝑑𝑑𝑑 = 3200 − 320 = 2880 ft⋅lbs
10
The total work done in pulling the container with the water to the top of the
building is therefore 320 ft⋅lbs+2880 ft⋅lbs=3200 ft⋅lbs.
(b) For the rope: In this case, we pull up the container with the water using the
rope but we do not neglect the weight of the rope. So, since the rope weighs
0.4 lbs/ft, then Δ𝑥𝑥 feet of rope weighs 0.4 Δ𝑥𝑥 pounds. Thus, at position 𝑥𝑥, the
rope weighs 𝐹𝐹 (𝑥𝑥 ) = 0.4 𝑥𝑥 pounds, and so the work done is
𝑊𝑊𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
1 2 80
= � 0.4 𝑥𝑥 𝑑𝑑𝑑𝑑 = � 𝑥𝑥 � = 16(80) = 1280 ft⋅lbs
5
0
0
80
Consequently, the total work required is given as
𝑊𝑊𝑊𝑊𝑊𝑊𝑘𝑘𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = 1280 + 3200 = 4480 ft⋅lbs
Problem 6. A chain lying on the ground is 10 meters long and its mass is 80 kg.
How much work is required to raise one end of the chain to a height of 6 meters?
Solution:
0
𝑥𝑥
6 − 𝑥𝑥
4
We have to make some assumptions about what is going on:
1. After lifting the one end; the chain is in an L shape, with 4 meters of it lying
on the ground.
2. The chain is able to slide along the ground with little or no effort while one
end is being lifted.
3. The weight density of the chain does not change throughout the portion of
the chain being lifted.
Let Δ𝑥𝑥 denote length of a piece of the chain which is 𝑥𝑥 meters below the raised
end of the chain. This same piece of chain must weigh exactly 78.4 Δ𝑥𝑥 Newtons;
and so the work done is Δ𝑊𝑊 = 78.4(6 − 𝑥𝑥 )Δ𝑥𝑥 Joules where 0 ≤ 𝑥𝑥 ≤ 6. Then
the total work requires is therefore,
1 2 6
� 78.4(6 − 𝑥𝑥 )𝑑𝑑𝑑𝑑 = 78.4 �6𝑥𝑥 − 𝑥𝑥 � = 78.4(18) = 1411.2 Joules
2
0
0
6
Problem 7. A uniform chain 10 feet long and weights 25 pounds is hanging from
a ceiling. Find the work required to lift the lower end of the chain to the ceiling so
that it is level with the upper end.
x
5
10
5
Solution:
The weight density the uniform chain is
25
10
= 2.5 lb/ft. The portion of the chain
that is at position 𝑥𝑥 below the ceiling for 5 ≤ 𝑥𝑥 ≤ 10 has to be lifted 2(5 − 𝑥𝑥)
feet. Therefore, the amount of work required for this portion will be 2(5 −
𝑥𝑥𝑖𝑖 )(2.5 Δ𝑥𝑥 ) and so the total amount of work will be
�
10
5
10
2(5 − 𝑥𝑥 )(2.5)𝑑𝑑𝑑𝑑 = 5 � (5 − 𝑥𝑥 ) 𝑑𝑑𝑑𝑑 = 62.5 ft⋅lbs
5
PUMPING LIQUIDS FROM CONTAINERS.
Question: How much work does it take to pump all or part of the liquid from a
container?
To find out, we imagine lifting the liquid out one thin horizontal slab at a time and
applying the formula (1) 𝑊𝑊 = 𝐹𝐹 ⋅ 𝑑𝑑 to each slab. We then evaluate the integral
we get each time depends on the weight of the liquid and the dimensions of the
container, but the way we find the integral is always the same as the next few
examples shows us what to do.
Problem 8. How much work does it take to pump water from a full upright
circular cylinder cylindrical tank of radius 5 meters and height 10 meters to a level
of 4 meters above the top of the tank?
Solution:
4m
10m
x
The gravitational force of magnitude 𝐹𝐹(𝑥𝑥) acting on the very thin disk of water
10 − 𝑥𝑥 meters below the top of the container is equal to the weight of that slab.
force = mass × gravity
= 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 × 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 × 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉
= (1000 kg/𝑚𝑚3 )(9.8 m/𝑠𝑠 2 ) × 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉
= 9800N/𝑚𝑚3 × 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉
Therefore the weight of the slab at elevation 𝑥𝑥 is
𝐹𝐹 = 𝐹𝐹 (𝑥𝑥 ) = (9800 N/𝑚𝑚3 )Δ𝑉𝑉 = (9800)(25𝜋𝜋)Δ𝑥𝑥 Newtons
Where Δ𝑉𝑉 = 𝜋𝜋(5 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 )2 (Δx meters) = 25𝜋𝜋Δ𝑥𝑥 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚. Thus the work
required to pump this slab of water 4 meters above the top of the contain is Δ𝑊𝑊
given as
Δi 𝑊𝑊 = Force × distance = (9800)(25𝜋𝜋Δ𝑥𝑥 ) ⋅ (10 − 𝑥𝑥 + 4) Joules
where 10 − 𝑥𝑥 + 4 meters is the distance the slab moves.Therefore, the total
work require to pump all the water 4 meter above the container is 𝑊𝑊 given by the
definite integral
𝑛𝑛
𝑛𝑛
𝑖𝑖=1
𝑖𝑖=1
𝑊𝑊 ≈ � Δ𝑖𝑖 𝑊𝑊 = � 24500𝜋𝜋 (14 − 𝑥𝑥𝑖𝑖 )Δ𝑥𝑥
1 2 10
𝑊𝑊ork = � 24500𝜋𝜋(14 − 𝑥𝑥 ) 𝑑𝑑𝑑𝑑 = 245000𝜋𝜋 �14𝑥𝑥 − 𝑥𝑥 �
2
0
0
= 245000𝜋𝜋 (140 − 50) = 22050000𝜋𝜋 ≈ 6.93 × 105 J 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜
10
A 1-horsepower output motor rate at 1 hp=746 J/s=746 W could empty the
tank is very little under 26 hours.
Problem 9. A tank in the form of an inverted right-circular cone whose planar
cross-section is bounded above by the line y=10, below by the 𝑥𝑥-axis, and on the
sides by the lines 𝑦𝑦 = 2𝑥𝑥 and 𝑦𝑦 = −2𝑥𝑥 is filled to within 2 feet of the top with
olive oil weighing 57 pounds per cubic feet. How much work does it take to pump
the oil to the rim of the tank?
Solution:
Top of the tank
𝑦𝑦 = 2𝑥𝑥
𝑦𝑦 = −2𝑥𝑥
Thin slab of olive oil
Oil level 𝑦𝑦 = 8
By similar triangles,
4
8
x
y
1
𝑥𝑥 4 1
= = ⇔ 𝑥𝑥 = 𝑦𝑦
2
𝑦𝑦 8 2
The magnitude of the force required to lift this slab is equal to is weight
(57 lbs/𝑓𝑓𝑡𝑡 3 for Olive oil),
𝐹𝐹(𝑦𝑦) = 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤ℎ𝑡𝑡 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 × 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉
1 2
= 57 𝜋𝜋 � 𝑦𝑦� Δ𝑦𝑦
2
57 2
=
𝜋𝜋𝜋𝜋 Δ𝑦𝑦 pounds
4
The distance through which 𝐹𝐹(𝑦𝑦) must act to lift this slab to the top of the tank is
𝑑𝑑 = (10 − 𝑦𝑦) feet, so the work done lifting the slab is
57 2
Δ𝑊𝑊 = Force ×distance =
𝜋𝜋𝑦𝑦 (10 − 𝑦𝑦)Δ𝑦𝑦 ft-lbs
4
The work done in pumping all the Olive oil to the top of the tank is the integral of
a diiferntial slab from 𝑦𝑦 = 0 to 𝑦𝑦 = 8 is
57 10 3 1 4 8
𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = �
𝜋𝜋𝑦𝑦 10 − 𝑦𝑦)𝑑𝑑𝑑𝑑 =
𝜋𝜋 � 𝑦𝑦 − 𝑦𝑦 � = 9728𝜋𝜋
4
3
4
4
0
0
≈ 30561 ft⋅lbs
8 57
2(
Problem 10. When gas expands in a cylinder with radius 𝑅𝑅, the pressure at any
given time is a function of the volume 𝑃𝑃 = 𝑃𝑃(𝑣𝑣). The force exerted by the gas on
the piston is the product of the pressure and the area : 𝐹𝐹 = 𝐹𝐹 (𝑣𝑣 ) = 𝜋𝜋𝑅𝑅 2 𝑃𝑃(𝑣𝑣 )
Show that the work done by the gas when the volume expands from volume 𝑣𝑣1 to
volume 𝑣𝑣2 is
𝑣𝑣2
𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = � 𝑃𝑃 𝑑𝑑𝑑𝑑 = �
𝑣𝑣1
𝑣𝑣=𝑣𝑣2
𝑣𝑣=𝑣𝑣1
𝑃𝑃 (𝑣𝑣 ) 𝑑𝑑𝑑𝑑
GAS
Force at position x
x
Solution: At time t, the piston is at position 𝑥𝑥 and so the volume of the gas at this
position must be 𝑣𝑣 = 𝑣𝑣 (𝑥𝑥 ) = 𝜋𝜋𝑅𝑅 2 𝑥𝑥 (we have a cylindrical container). The force
on the piston at position 𝑥𝑥 must then be
𝐹𝐹 (𝑣𝑣(𝑥𝑥)) = 𝜋𝜋𝑅𝑅 2 𝑃𝑃�𝑣𝑣 (𝑥𝑥 )� = 𝜋𝜋𝑅𝑅 2 𝑃𝑃.
Thus, if the piston changes from position 𝑥𝑥 = 𝑥𝑥1 to position 𝑥𝑥 = 𝑥𝑥2 then the
volume changes from 𝑣𝑣 = 𝜋𝜋𝑅𝑅 2 𝑥𝑥1 = 𝑣𝑣1 to 𝑣𝑣 = 𝜋𝜋𝑅𝑅 2 𝑥𝑥2 = 𝑣𝑣2 . The work done on
the piston during small change Δ𝑥𝑥 in position is
Δi 𝑊𝑊 = Force×Distance= 𝜋𝜋𝑅𝑅 2 Pi Δ𝑥𝑥 = 𝐹𝐹𝑖𝑖 Δ𝑥𝑥 = 𝐹𝐹𝑖𝑖 ⋅ �
1
𝜋𝜋𝑅𝑅 2
Δ𝑣𝑣� = 𝑃𝑃𝑖𝑖 Δ𝑣𝑣 ,
where 𝑥𝑥1 < 𝑥𝑥𝑖𝑖−1 ≤ 𝜉𝜉𝑖𝑖 ≤ 𝑥𝑥𝑖𝑖 < 𝑥𝑥2 and 𝑣𝑣1 < 𝑣𝑣𝑖𝑖−1 ≤ 𝑣𝑣𝑖𝑖∗ ≤ 𝑣𝑣𝑖𝑖 < 𝑣𝑣2 The total work
done of the piston by gas is then nearly,
𝑛𝑛
𝑛𝑛
𝑛𝑛
𝑖𝑖=1
𝑖𝑖=1
𝑖𝑖=1
1
Total Work done ≈ � 𝐹𝐹𝑖𝑖 (𝑣𝑣𝑖𝑖∗) Δ𝑥𝑥 = � 𝐹𝐹𝑖𝑖 (𝑣𝑣𝑖𝑖∗) 2 Δ𝑣𝑣 = � 𝑃𝑃𝑖𝑖 Δ𝑣𝑣
πR
Total Work done =
lim
max Δ𝑣𝑣→0
𝑛𝑛
� 𝑃𝑃𝑖𝑖 Δ𝑣𝑣 = �
𝑖𝑖=1
𝑣𝑣=𝑣𝑣2
𝑣𝑣=𝑣𝑣1
𝑣𝑣2
𝑃𝑃(𝑣𝑣 )𝑑𝑑𝑑𝑑 = � 𝑃𝑃 𝑑𝑑𝑑𝑑
𝑣𝑣1
HOOKE’S LAW FOR SPINGS
In the next example we use Hooke’s Law, which states that:
under appropriate conditions if a linear spring is elongated 𝑥𝑥 inches beyond its
natural (unstretched) length, then the magnitude 𝐹𝐹 of the “restoring force” (the
pull-back force) is proportional to the elongation 𝑥𝑥, where 𝑘𝑘 (called the force
constant orthe spring constant) is a constant (with British units lb/ft or metric
units N/m) depending on materials used to construct the spring. That is,
𝐹𝐹 = 𝑘𝑘𝑘𝑘
x
Hooke’s Law gives good results provided that the force does not distort the
material of which the spring is made. We assume that the forces in this discussion
are too small to do that.
Thus, the work done by the force when the elongation goes from 𝑥𝑥 = 0 to a
maximum value 𝑥𝑥 = 𝑋𝑋 is
𝑋𝑋
𝑋𝑋
1 1
1
⏟ = 𝑘𝑘𝑋𝑋 2
𝑊𝑊𝑜𝑜𝑜𝑜𝑜𝑜 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 = � 𝐹𝐹(𝑥𝑥) 𝑑𝑑𝑑𝑑 = � 𝑘𝑘𝑘𝑘 𝑑𝑑𝑑𝑑 = � ⋅ 𝑘𝑘𝑋𝑋 2 � ⋅ 𝑋𝑋
�������
𝑋𝑋 2
2
0
0
distance
average force
Problem 11: A spring has a natural length of 14 meters . If a force of 5 N is
required to keep the spring stretched 2 meters. How much work is done in
stretching the spring from its natural length to a length of 18 meters.
Solution:
x
0
4
We must first find the spring constant 𝑘𝑘. By Hooke’s Law, 𝐹𝐹 (𝑥𝑥 ) = 𝑘𝑘𝑘𝑘. Thus, 𝐹𝐹(2) =
𝑘𝑘(2) = 5 and so we can calculate 𝑘𝑘 as follows
𝑘𝑘 =
𝐹𝐹(2) 5 𝑁𝑁
=
= 2.5 𝑁𝑁/𝑚𝑚
2
2 𝑚𝑚
Therefore, 𝐹𝐹(𝑥𝑥 ) = 2.5𝑥𝑥 and so the work done by the force is
5 2 4
Work done = � 2.5 𝑥𝑥 𝑑𝑑𝑑𝑑 = � 𝑥𝑥 � = 20 Joules
4
0
0
4
WORK AND KINETIC ENERGY RELATIONSHIP
If a variable force of magnitude 𝐹𝐹(𝑥𝑥) moves a body of mass 𝑀𝑀 along the 𝑥𝑥-axis
from position 𝑥𝑥1 to position 𝑥𝑥2 , the body’s speed 𝑣𝑣 can be written as 𝑑𝑑𝑑𝑑/𝑑𝑑𝑑𝑑
(where 𝑡𝑡 represent time).
Using Newton’s Second Law of Motion:
𝐹𝐹 = 𝑀𝑀𝑀𝑀 = 𝑀𝑀
𝑑𝑑𝑑𝑑
𝑑𝑑𝑥𝑥 𝑑𝑑𝑣𝑣
𝑑𝑑𝑑𝑑
= 𝑀𝑀 � ⋅ � = 𝑀𝑀 𝑣𝑣
𝑑𝑑𝑑𝑑
𝑑𝑑𝑡𝑡 𝑑𝑑𝑥𝑥
𝑑𝑑𝑑𝑑
to show that the net work done by the force in moving the body from 𝑥𝑥1 to 𝑥𝑥2 is
𝑥𝑥 2
𝑊𝑊 = � 𝐹𝐹 (𝑥𝑥 ) 𝑑𝑑𝑑𝑑
𝑥𝑥 1
𝑥𝑥 2
𝑑𝑑𝑑𝑑
1 2 𝑥𝑥 2 1
1
= � 𝑀𝑀𝑀𝑀
𝑑𝑑𝑑𝑑 = � 𝑀𝑀𝑀𝑀 𝑑𝑑𝑑𝑑 = 𝑀𝑀 � 𝑣𝑣 � = 𝑀𝑀𝑣𝑣22 − 𝑀𝑀𝑣𝑣12
𝑑𝑑𝑑𝑑
2
2
2
𝑥𝑥 1
𝑥𝑥 1
𝑥𝑥 1
Where 𝑣𝑣1 and 𝑣𝑣2 are the body’s speed at positions 𝑥𝑥1 and 𝑥𝑥2 . In physics, the
expression
1
𝑀𝑀𝑣𝑣 2
2
is called the kinetic energy (K) of a body of mass 𝑀𝑀 moving with speed 𝑣𝑣.
Therefore, the work done by the force equals the change in the body’s kinetic
energy, and we can find work by calculating this change. That is,
𝑥𝑥 2
𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = Δ 𝐾𝐾 = 𝑐𝑐ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑖𝑖𝑛𝑛 𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸
Problem 12. During the match in which Pete Sampras won the 1990 U.S. Open
men’s tennis championship, Sampras hit a serve that was clocked at a
phenomenal 124 mph. How much work did Sampras have to do on the 2 oz ball
to get to that speed?
Solution:
2
𝑀𝑀 =
oz = (0.125 lbs)/(32 ft/s 2 ) and
32
Thus, work is
𝑣𝑣 = 124 mph = 124 mi/hr × 5280 ft/mi ×
1
hr/s
3600
1 0.125
5280
𝑊𝑊𝑜𝑜𝑜𝑜𝑜𝑜 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 = �
� �124 ⋅
� = 0.355 ft⋅lbs
2 32
3600