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 Work done by a spring
 Work done by pumping a liquid
Work done by a
spring
Hooke’s Law states that within the limits of
elasticity the displacement produced in a body is
proportional to the force applied, that is,
F = kx,
where the constant k is the constant of
proportionality called the modulus.
Thus , F(x) = kx
The work done is
HOOKE’S LAW
1. If the modulus of a spring is 20 lbs./in., what
is the work required to stretch the spring a
distance of 6 inches?
2. If a force of 50 lbs. stretches a 12 in.
spring to 14 in., find the work done in
stretching the spring from 15 in. to 17 in.
3. A spring has a natural length of 10 inches. An 800-lb
force stretches the spring 14-inches. (a) Find the force
constant. (b) How much work is done in stretching the
spring from 10 inches to 12 inches? (c) How far beyond its
natural length will a 1600-lb. force stretch the spring?
3. A spring has a natural length of 10 inches. An 800-lb force
stretches the spring 14-inches. (a) Find the force constant. (b)
How much work is done in stretching the spring from 10 inches
to 12 inches? (c) How far beyond its natural length will a 1600-
lb. force stretch the spring?
4. A force of 200 N will stretch a garage door spring 0.8-
m beyond its unstressed length. How far will a 300-N-
force stretch the spring? How much work does it take to
stretch the spring this far?
SOLUTION:
To determine how far a 300-N-force will stretch the spring, we
must first determine the force constant using F= kx.
200 = 0.8k k = 250 N /m
Thus , F= 250x
300 = 250x x = 1.2 m
To determine the work done to stretch the spring
this far,
JmNxxdxW 180180]125250 2.1
0
2
2.1
0
=−=== ∫
4. A crate is pushed a distance of 15 meters. If it is
pushed with a force equvalent to 4x + 10 newtons, how
much work was done to move the crate?
SOLUTION:
∫=
b
a
dxxFW )(
104)( += xxF
600]102 15
0
2
=+= xxW Joules
5. A force of 1200 N compresses a spring from its natural
length of 18 cm to a length of 16 cm. How much work is
done in compressing it from 16 cm to 14 cm?
SOLUTION:
F = kx
1200 = k(2)
k = 600 N /cm Thus F (x)= 600x
[ ]
m-NW
cmNW
xW
xdxW
36
3600
300
600
4
2
2
4
2
=
−=
=
= ∫
Work done by
pumping
Work done in Pumping a
Liquid
The total work done in lifting all or part of the
liquid in a container to any point P above its
top is
where w = weight per unit volume of the liquid
h = distance of the element from the
point P
dv = volume of the solid generated by
revolving the element
∫
∫
=
=
b
a
b
a
hdVwW
whdVW
EXAMPLE
1. A swimming pool full of water is in the form of a rectangular
parallelepiped 5 m deep, 25 m long and 15 m wide. Find the
work required to pump the water in the pool up to a level one
meter above the surface of the pool.
∫=
b
a
hdVwW
lwhV =
for the element of the volume,
dydV )25)(15(=
using
∫ −=
5
0
)375)(6( dyywW






−=
2
6375
2
y
ywW
0
5
wW
2
13125
= dyne-m
5
25
15
6
h= 6-y
y
EXAMPLE
2. The inner surface of a tank has the form of a parabola of
revolution whose axis is vertical. The depth of the tank and
the diameter of the circular top are 12 cm. If the tank is
originally full of water, find the work done in pumping all the
water:
a. To the top
b. 3 cm from the top
c. Suppose the tank is half-full in (a)
r =6
12
y
h= 12 - y
(6,12)
x
y
for the element of the volume, (strip is in the form of a cylinder) thus
hrV 2
π=
dyxdV 2
π=
to find the equation of the parabola, we use and substitute the
coordinates of the point (6,12) to find 4a.
,42
ayx =
)12(462
a= 34 =a
Thus , substitute inyx 32
= dyxdV 2
π=
dyydV )3(π=
( )
wW
ydyywW
hdVwW
π=
π−=
=
∫
∫
864
312
12
0
12
0
dyne-cm
a.
b. If the water is to be pumped 3 cm above its surface, the only value which will
change is h; h = 15-y
cmdyneW
ydyywW
hdVwW
−=
−=
=
∫
∫
_______
3)15(
12
0
12
0
π
Thus
c. If the tank is half-full, just change the limit of (a) from 0 to 6 since the
container is half-full.
1. A conical vessel full of water is 16 ft. across the top and 12 ft.
deep. Find the work required to pump all the water to a point 2
ft above the top of the vessel.
2. A tank is in the shape of a right circular cone with height 5 m and
top radius 2 m. It contains water up to the height of 4 m. The
density of water is 1,000 kg/m3
. How much work must be done to
pump all of the water out of the tank over the top edge of the
tank?
3. Suppose that a cylindrical tank has height 10, the radius of the
base is 7, and it is half filled with water. Find the amount of work
necessary to move all of the water out of the top of the tank.
EXERCISES

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Lesson 17 work done by a spring and pump final (1)

  • 1.  Work done by a spring  Work done by pumping a liquid
  • 2. Work done by a spring
  • 3. Hooke’s Law states that within the limits of elasticity the displacement produced in a body is proportional to the force applied, that is, F = kx, where the constant k is the constant of proportionality called the modulus. Thus , F(x) = kx The work done is HOOKE’S LAW
  • 4. 1. If the modulus of a spring is 20 lbs./in., what is the work required to stretch the spring a distance of 6 inches? 2. If a force of 50 lbs. stretches a 12 in. spring to 14 in., find the work done in stretching the spring from 15 in. to 17 in. 3. A spring has a natural length of 10 inches. An 800-lb force stretches the spring 14-inches. (a) Find the force constant. (b) How much work is done in stretching the spring from 10 inches to 12 inches? (c) How far beyond its natural length will a 1600-lb. force stretch the spring?
  • 5. 3. A spring has a natural length of 10 inches. An 800-lb force stretches the spring 14-inches. (a) Find the force constant. (b) How much work is done in stretching the spring from 10 inches to 12 inches? (c) How far beyond its natural length will a 1600- lb. force stretch the spring?
  • 6. 4. A force of 200 N will stretch a garage door spring 0.8- m beyond its unstressed length. How far will a 300-N- force stretch the spring? How much work does it take to stretch the spring this far? SOLUTION: To determine how far a 300-N-force will stretch the spring, we must first determine the force constant using F= kx. 200 = 0.8k k = 250 N /m Thus , F= 250x 300 = 250x x = 1.2 m
  • 7. To determine the work done to stretch the spring this far, JmNxxdxW 180180]125250 2.1 0 2 2.1 0 =−=== ∫
  • 8. 4. A crate is pushed a distance of 15 meters. If it is pushed with a force equvalent to 4x + 10 newtons, how much work was done to move the crate? SOLUTION: ∫= b a dxxFW )( 104)( += xxF 600]102 15 0 2 =+= xxW Joules
  • 9. 5. A force of 1200 N compresses a spring from its natural length of 18 cm to a length of 16 cm. How much work is done in compressing it from 16 cm to 14 cm? SOLUTION: F = kx 1200 = k(2) k = 600 N /cm Thus F (x)= 600x [ ] m-NW cmNW xW xdxW 36 3600 300 600 4 2 2 4 2 = −= = = ∫
  • 11. Work done in Pumping a Liquid The total work done in lifting all or part of the liquid in a container to any point P above its top is where w = weight per unit volume of the liquid h = distance of the element from the point P dv = volume of the solid generated by revolving the element ∫ ∫ = = b a b a hdVwW whdVW
  • 12. EXAMPLE 1. A swimming pool full of water is in the form of a rectangular parallelepiped 5 m deep, 25 m long and 15 m wide. Find the work required to pump the water in the pool up to a level one meter above the surface of the pool. ∫= b a hdVwW lwhV = for the element of the volume, dydV )25)(15(= using ∫ −= 5 0 )375)(6( dyywW       −= 2 6375 2 y ywW 0 5 wW 2 13125 = dyne-m
  • 14. EXAMPLE 2. The inner surface of a tank has the form of a parabola of revolution whose axis is vertical. The depth of the tank and the diameter of the circular top are 12 cm. If the tank is originally full of water, find the work done in pumping all the water: a. To the top b. 3 cm from the top c. Suppose the tank is half-full in (a)
  • 15. r =6 12 y h= 12 - y (6,12) x y for the element of the volume, (strip is in the form of a cylinder) thus hrV 2 π= dyxdV 2 π= to find the equation of the parabola, we use and substitute the coordinates of the point (6,12) to find 4a. ,42 ayx = )12(462 a= 34 =a
  • 16. Thus , substitute inyx 32 = dyxdV 2 π= dyydV )3(π= ( ) wW ydyywW hdVwW π= π−= = ∫ ∫ 864 312 12 0 12 0 dyne-cm a. b. If the water is to be pumped 3 cm above its surface, the only value which will change is h; h = 15-y cmdyneW ydyywW hdVwW −= −= = ∫ ∫ _______ 3)15( 12 0 12 0 π Thus
  • 17. c. If the tank is half-full, just change the limit of (a) from 0 to 6 since the container is half-full.
  • 18. 1. A conical vessel full of water is 16 ft. across the top and 12 ft. deep. Find the work required to pump all the water to a point 2 ft above the top of the vessel. 2. A tank is in the shape of a right circular cone with height 5 m and top radius 2 m. It contains water up to the height of 4 m. The density of water is 1,000 kg/m3 . How much work must be done to pump all of the water out of the tank over the top edge of the tank? 3. Suppose that a cylindrical tank has height 10, the radius of the base is 7, and it is half filled with water. Find the amount of work necessary to move all of the water out of the top of the tank. EXERCISES