6.4: Work
(Dated: October 7, 2011)
Let x(t ) be the position function of an object of mass m
moving along a straight line as a function of time t . Then
the force f on the object can be calculated from Newton’s
second law:
f (t ) = m
d 2x
.
dt2
In the SI metric system, the units for the mass, the displacement, the time, and the force are respectively kilograms (kg),
meters (m), seconds (s), and newtons (N). Hence Newton’s
second law says that a force of 1 N acting on a mass of 1 kg
induces an acceleration of 1 m/s2 .
If the force is measured in pounds and the distance in
feet, then the unit for the work is a foot-pount (ft-lb).
The work W done by a constant force f to move an object
a distance x is given by the product
W = f x.
In SI metric system, the work is measured in joules (J). In
other words, 1 J= 1 N−m.
The work W done by a continuous force f to move an object from x = a to x = b can be approximated by a Riemann
sum and calculated as an integral. To see this, let’s divide the
b−a
interval [a, b] into n subintervals of equal width ∆x =
.
n
Then the work Wi done by the force f to move the object
from x i −1 to x i , where [x i −1 , x i ] is the i th subinterval, can
be approximated by
Wi = f (x i∗ )∆x,
where x i∗ is a sample point in [x i −1 , x i ]. This approximation
gets better as one decreases the width ∆x (by increasing the
number of subintervals n). This is so because f is continuous and thus doesn’t change much over sufficiently small
interval. In consequence, the work W done by the force f in
moving the object from x = a to x = b satisfies the condition
W≈
n
X
i =1
Wi =
n
X
i =1
n→∞
n
X
i =1
Example 3 (Exercise 6.4.9** in the text) Hooke’s law states
that the force f (x) required to maintain a spring stretched
x units beyond its natural length is proportional to x:
Hooke’s law: f (x) = kx.
Here k is a positive proportional constant. It is usually called
the spring constant. This is equivalent to saying that the work
needed to maintain the spring stretched x units beyond its
natural length is
f (x i∗ )∆x.
f (x i∗ )∆x =
Z
b
f (x) d x.
a
Example 1 (Exercise 6.4.3* in the tex) A particle is moved
10
along the x-axis by a force that measures f (x) =
(1 + x)2
pounds at a point x feet from the origin. Find the work done
W in moving the particle from the origin to a distance of 9 ft.
Solution.
Z 9
Z 10
10
du
u=1+x
W=
d
x
=
10
= 9 ft-lb.
2
u2
0 (1 + x)
1
x
Z
Hooke’s law: W =
0
k x̃ d x̃ =
kx 2
.
2
Suppose that 2 J of work is needed to stretch a spring from its
natural length of 30 cm to a length of 42 cm.
(a) How much work is needed to stretch the spring from 35
cm to 40 cm?
(b) How far beyond its natural length will a force of 30 N
keep the spring stretched?
Solution. By Hooke’s law,
k(42 cm − 30 cm)2
= 2 J.
2
1
2500
2500
Thus k =
J-cm−2 =
J-m−2 =
N-m−1 . (Note: 1
36
9
9
m= 100 cm.)
(a) The work needed to stretch the spring from 35 cm to 40
cm is
Z (40−30) cm
Z (35−30) cm
W=
kx d x −
kx d x
0
0
Z 10 cm
=k
x dx
5
x2
=k
2
·
As one passes to the limit n → ∞, one gets
W = lim
Example 2 (Exercise 6.4.5 in the text) Refer to the text.
Hint: Area under the graph. Ans: 180 J.
¸10 cm
5
·
¸
(10 cm)2 (5 cm)2
1
=
J-cm−2
−
36
2
2
25
J
=
24
≈ 1.04167 J.
(b) By Hooke;s law,
30 N = kx =
2500
N-m−1 · x,
9
where x is the elongation of the string from its natural length
due to the force of 30 N. Thus
x=
9
27
· 30 =
≈ 0.108 m = 10.8 cm.
2500
250
2
Example 4 (Exercise 6.4.11 in the text) A spring has natural length 20 cm. Compare the work W1 done in stretching
the spring from 20 cm to 30 cm with the work W2 done in
stretching it from 30 cm to 40 cm. How are W1 and W2 related?
Solution. By Hooke’s law,
Z (30−20) cm
k £ 2 ¤10 cm
x 0
= 50k cm2 ,
W1 =
kx d x =
2
0
while
Z
W2 =
(40−20) cm
kx d x =
30−20
k £ 2 ¤20 cm
x 10 = 150k cm2 ,
2
where k is the spring constant. Thus W2 = 3W1 .
Example 5 (Exercise 6.4.12* in the text) If 6 J of work is
needed to stretch a spring from 10 cm to 12 cm and another
10 J is needed to stretch it from 12 cm to 14 cm, what is the
natural length of the spring?
Solution. Say L cm is the natural length of the spring. By
Hooke’s law,
Z (12−L cm
¤
k£
6J=
kx d x =
(12 − L)2 − (10 − L)2 cm2
2
10−L
Z
10 J =
(14−L cm
12−L
kx d x =
¤
k£
(14 − L)2 − (12 − L)2 cm2 .
2
Dividing the second equation by the first yields
5 (14 − L)2 − (12 − L)2
=
.
3 (12 − L)2 − (10 − L)2
On the other hand, the work done in lifting the coal is
Wcoal = 800 × 500 = 400000 ft-lb.
Hence the total work done is
W = Wcable + Wcoal = 650000 ft-lb.
Example 7 (Exercise 6.4.18* in the text) A
10-ft
chain
weighs 25 lb and hangs from a ceiling. Find the work done
in lifting the lower end of the chain to the ceiling so that it’s
level with the upper end.
Solution. Let’s place the origin in the middle of the chain
and the x-axis pointing downward. This way the lower end
is initially at x = 5 ft. Dividing the interval [0, 5] into n
10 − 5
subtervals of equal length ∆x =
divides the lower half
n
25
of the chain into n parts, each weighs
∆x = 2.5∆x lb. It
10
follows that the work done in lifting the i th part of the chain
from x i∗ to −x i∗ is given by
Wi ≈ 2.5∆x × 2x i∗ = 5x i∗ ∆x in lb-ft,
where x i∗ is a sample point in the i th subinterval [x i −1 , x i ].
Note that lifting the lower end of the chain to the ceiling so
that it’s level with the upper end is equivalent to lifting the
i th part of the chain from x i∗ to −x i∗ for all i , i = 1, 2, · · · , n.
Thus the total work in lifting the lower end of the chain to the
ceiling so that it’s level with the upper end is
W = lim
n→∞
Simplifying this equation gives 8L − 64 = 0 or L = 8. Thus the
natural length of the spring is L cm= 8 cm.
= lim
Example 6 (Exercise 6.4.15 in the text) A cable weighs 2 lbft−1 is used to lift 800 lb of coal up a mine shaft 500 ft deep.
Find the work done.
Solution. Let’s place the origin at the top of the mine shaft
and the x-axis pointing downward. This way the bottom
of the mine shaft has a coordinate of x = 500 ft. Dividing the interval [0, 500] into n subintervals of equal length
500 − 0
∆x =
divides the cable into n parts, each weighs 2∆x
n
lb. Thus the work done in lifting the i th part to the top of the
mine shaft is given by
=
Wi ≈ 2x i∗ ∆x in lb-ft,
where x i∗ is a sample point in the i th subinterval [x i −1 , x i ].
Consequently, the work done in lifting the cable is
Wcable = lim
n→∞
= lim
n→∞
i =1
n
X
i =1
Wi
2x i∗ ∆x
500
Z
=
n
X
2x d x
0
= 250000 ft-lb.
n→∞
n
X
i =1
n
X
i =1
Wi
5x i∗ ∆x
5
Z
5x d x
0
= 62.5 ft-lb.
Example 8 (Exercise 6.4.21* in the text) A tank is full of
water. Find the work required to pump the water out of the
spout. (Refer to the text for a diagram of the tank. Also, take
the gravitational acceleration as g = 9.8 m-s−2 and the density (i.e., mass per unit volume) of water as 1000 kg-m−3 .)
Solution. Let’s measure the depths (measured in meters)
from the top of the spout by introducing a vertical coordinate
line. This way the coordinate of the spout is x = 0 units and
the coordinate of the bottom of the tank is x = 5. The water
extends from a depth of 2 units to a depth of 5 units.
Dividing the interval [2, 5] into n subintervals of equal
5−2
divides the water into n layers, each of
width ∆x =
n
which has a height of ∆x. Let [x i −1 , x i ] be the i th subinterval. The i th layer, i.e., the one lying between the depths of
x i −1 and x i is a rectangle of length 8 units and width approximately 5 − x i∗ units, where x i∗ is a sample point in [x i −1 , x i ].
(Convince yourself that this is the case.) Hence it has a volume of approximately
Vi ≈ 8(5 − x i∗ )∆x in m3 .
3
Hence the mass of the i th layer is approximately
m i = 1000Vi ≈ 8000(5 − x i∗ )∆x in kg.
The force required to raise this layer must overcome the gravitational force and so is given by
F i = m i g ≈ 78400(5 − x i∗ )∆x in N.
Dividing the interval [0, 8] into n subintervals of equal
8−0
divides the water into n layers, each of
width ∆x =
n
which has a height of ∆x. Let [x i −1 , x i ] be the i th subinterval.
The i th layer, i.e., the one lying between the depths of x i −1
48 − 3x i∗
and x i is a circular cylinder of radius r i =
units,
8
∗
where x i is a sample point in [x i −1 , x i ]. (Convince yourself
that this is the case.) Hence it has a volume of approximately
It follows that the work Wi done in raising this layer to the
spout is approximately
Wi = F i x i∗ ≈ 78400(5 − x i∗ )x i∗ ∆x in J.
Consequently, the total work done in emptying the tank is
approximately
W≈
n
X
n
X
Wi =
i =1
i =1
78400(5 − x i∗ )x i∗ ∆x in J,
the exact value of which is obtained by passing to the limit
n → ∞:
W = lim
n→∞
i =1
78400(5 − x i∗ )x i∗ ∆x
2
78400(5 − x)x d x
= 1058400 J.
µ
48 − 3x i∗
¶2
∆x in ft3 .
8
Hence the force required to raise this layer is given by
µ
F i = 62.5 × Vi ≈ 62.5π
48 − 3x i∗
¶2
8
∆x in lb.
Note that this is equivalent to the weight of the i th layer. Consequently, the work Wi done in raising this layer to the spout
is approximately
Wi = F i x i∗
≈ 62.5Vi ≈ 62.5πx i∗
µ
48 − 3x i∗
¶2
∆x in lb-ft.
8
The total work done in emptying the tanks is approximated by
5
Z
=
n
X
Vi ≈ π
W=
n
X
Wi =
i =1
n
X
i =1
62.5πx i∗
µ
48 − 3x i∗
¶2
8
∆x in lb-ft.
Passing to the limit n → ∞ yields the exact value
Example 9 (Exercise 6.4.23 in the text) A tank is full of water. Find the work required to pump the water out of the
spout. (Refer to the text for a diagram of the tank. Also, use
the fact that water weighs 62.5 lb per ft.)
Solution. Let’s measure the depths (measured in feet) from
the top of the spout by introducing a vertical coordinate line.
This way the coordinate of the spout is x = 0 units and the
coordinate of the bottom of the tank is x = 8.
W = lim
n→∞
i =1
62.5πx i∗
8
Z
=
n
X
µ
62.5πx
0
µ
48 − 3x i∗
48 − 3x
8
= 103673 lb-ft.
8
¶2
dx
¶2
∆x