# Lesson 22 Work Done by A Apring and Pump2 ``` Work done by a spring
 Work done by pumping a liquid
Work done by a spring
Hooke’s Law
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 = kx
The work done is
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?
F(x) = 20x
W = ∫60 20xdx
6
= &frac12; 20 x ]0
= 10(6)2 - 10(0)2
= 10 (36)
W = 360 lbs. in.
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.
To get the limits of integration
15-12 = 3
x=0
x=3
x=5
17-12 = 5
if x = 2, F = 50
F = kx
50 = 2k
K = 25
F(x)= 25x
12 in
W = ∫5 25xdx
3
5
2
= &frac12; 25x ]3
= 25(5)2 /2 - 25(3) 2 /2
= 625/2 - 225/2
W = 200 lbs.-in.
15in
17in
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?
SOLUTION:
PART (a)
To find the force constant (spring constant), we first solve for
k. Substitute 800 in. for F and (14 - 10) in. for x.
F = kx
800 = k(14 - 10)
k = 800/4 = 200 lbs./ft
PART (b)
Thus F(x) = 200x; to calculate the work done in
stretching the spring from 10 inches to 12 inches,
b
use
W   F ( x)dx
a
2
W   200xdx  100x 2 ]02  400in  lbs
0
PART (c)
To determine how far 1600-lb. force will stretch the
spring, we do not need to integrate.
We use
F= 200x
1600 = 200x
x = 8 inches
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,
1.2
W   250xdx  125x 2 ]10.2  180N  m  180J
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:
F ( x)  4 x  10
b
W   F ( x)dx
a
W  2x  10x]  600 Joules
2
15
0
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
4
W   600xdx

2

2 4
2
W  300x
W  3600N  m
Thus F (x)= 600x
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
b
top is
W 
wh d V

a
b
W  w h d V
a
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
EXAMPLE
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.
for the element of the volume,
V  lwh
dV  (15)(25)dy
using
b
W  w hdV
a
5
W  w (6  y)(375)dy
0
5

y2 
W  375 w6 y  
2

0
13125
W
w dyne-m
2
6
h= 6-y
y
25
5
15
EXAMPLE
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)
y
(6,12)
r =6
h= 12 - y
12
y
x
4a
for the element of the volume, (strip is in the form of a cylinder) thus
V  r 2 h
dV  x 2 dy
to find the equation of the parabola, we use x  4ay, and substitute the
coordinates of the point (6,12) to find 4a. 62  4a(12)
4a  3
2
Thus
x2  3 y
dV  x 2 dy
, substitute in
dV   (3 y)dy
a.
12
W   hdV
0
W  w 12  y  3 ydy
12
0
W  864w dyne-cm
b. If the water is to be pumped 3 cm above its surface, the only value which will
change is h; h = 15-y
Thus
12
W  w hdV
0
12
W  w (15  y )3ydy
0
W  _______dyne cm
c. If the tank is half-full, just change the limit of (a) from 0 to 6 since the
container is half-full.
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