160
CHAPTE R 7 KI N ETIC E N E RGY AN D WOR K
state. For this common arrangement, we can write Eq. 7-20 as
F x ϭ Ϫkx (Hooke’s law),
(7-21)
where we have changed the subscript. If x is positive (the spring is stretched
toward the right on the x axis), then F x is negative (it is a pull toward the left). If
x is negative (the spring is compressed toward the left), then F x is positive (it is a
push toward the right). Note that a spring force is a variable force because it is a
function of x, the position of the free end.Thus F x can be symbolized as F(x). Also
note that Hooke’s law is a linear relationship between F x and x.
The Work Done by a Spring Force
To find the work done by the spring force as the block in Fig. 7-10a moves, let us
make two simplifying assumptions about the spring. (1) It is massless; that is, its
mass is negligible relative to the block’s mass. (2) It is an ideal spring; that is, it
obeys Hooke’s law exactly. Let us also assume that the contact between the block
and the floor is frictionless and that the block is particle-like.
We give the block a rightward jerk to get it moving and then leave it alone.
As the block moves rightward, the spring force F x does work on the block,
decreasing the kinetic energy and slowing the block. However, we cannot find this
work by using Eq. 7-7 (W ϭ Fd cos f) because there is no one value of F to plug
into that equation—the value of F increases as the block stretches the spring.
There is a neat way around this problem. (1) We break up the block’s displacement into tiny segments that are so small that we can neglect the variation
in F in each segment. (2) Then in each segment, the force has (approximately) a
single value and thus we can use Eq. 7-7 to find the work in that segment. (3)
Then we add up the work results for all the segments to get the total work. Well,
that is our intent, but we don’t really want to spend the next several days adding
up a great many results and, besides, they would be only approximations. Instead,
let’s make the segments infinitesimal so that the error in each work result goes to
zero. And then let’s add up all the results by integration instead of by hand.
Through the ease of calculus, we can do all this in minutes instead of days.
Let the block’s initial position be x i and its later position be x f . Then divide
the distance between those two positions into many segments, each of tiny length
⌬x. Label these segments, starting from x i , as segments 1, 2, and so on. As the
block moves through a segment, the spring force hardly varies because the segment is so short that x hardly varies. Thus, we can approximate the force magnitude as being constant within the segment. Label these magnitudes as F x1 in
segment 1, F x2 in segment 2, and so on.
With the force now constant in each segment, we can find the work done
within each segment by using Eq. 7-7. Here f ϭ 180Њ, and so cos f ϭ Ϫ1. Then
the work done is ϪF x1 ⌬x in segment 1, ϪF x2 ⌬x in segment 2, and so on. The net
work W s done by the spring, from x i to x f , is the sum of all these works:
(7-22)
where j labels the segments. In the limit as ⌬x goes to zero, Eq. 7-22 becomes
(7-23)
From Eq. 7-21, the force magnitude F x is kx.Thus, substitution leads to
(7-24)
ϭ (Ϫ
1
2 k)[x
2
] x i
x f ϭ (Ϫ
1
2 k)(x f
2 Ϫ x i
2
).
W s ϭ ͵
x f
x i
Ϫkx dx ϭ Ϫk ͵
x f
x i
x dx
W s ϭ ͵
x f
x i
ϪF x dx.
W s ϭ ͚ ϪF xj ⌬x,
CHAPTE R 7 KI N ETIC E N E RGY AN D WOR K
state. For this common arrangement, we can write Eq. 7-20 as
F x ϭ Ϫkx (Hooke’s law),
(7-21)
where we have changed the subscript. If x is positive (the spring is stretched
toward the right on the x axis), then F x is negative (it is a pull toward the left). If
x is negative (the spring is compressed toward the left), then F x is positive (it is a
push toward the right). Note that a spring force is a variable force because it is a
function of x, the position of the free end.Thus F x can be symbolized as F(x). Also
note that Hooke’s law is a linear relationship between F x and x.
The Work Done by a Spring Force
To find the work done by the spring force as the block in Fig. 7-10a moves, let us
make two simplifying assumptions about the spring. (1) It is massless; that is, its
mass is negligible relative to the block’s mass. (2) It is an ideal spring; that is, it
obeys Hooke’s law exactly. Let us also assume that the contact between the block
and the floor is frictionless and that the block is particle-like.
We give the block a rightward jerk to get it moving and then leave it alone.
As the block moves rightward, the spring force F x does work on the block,
decreasing the kinetic energy and slowing the block. However, we cannot find this
work by using Eq. 7-7 (W ϭ Fd cos f) because there is no one value of F to plug
into that equation—the value of F increases as the block stretches the spring.
There is a neat way around this problem. (1) We break up the block’s displacement into tiny segments that are so small that we can neglect the variation
in F in each segment. (2) Then in each segment, the force has (approximately) a
single value and thus we can use Eq. 7-7 to find the work in that segment. (3)
Then we add up the work results for all the segments to get the total work. Well,
that is our intent, but we don’t really want to spend the next several days adding
up a great many results and, besides, they would be only approximations. Instead,
let’s make the segments infinitesimal so that the error in each work result goes to
zero. And then let’s add up all the results by integration instead of by hand.
Through the ease of calculus, we can do all this in minutes instead of days.
Let the block’s initial position be x i and its later position be x f . Then divide
the distance between those two positions into many segments, each of tiny length
⌬x. Label these segments, starting from x i , as segments 1, 2, and so on. As the
block moves through a segment, the spring force hardly varies because the segment is so short that x hardly varies. Thus, we can approximate the force magnitude as being constant within the segment. Label these magnitudes as F x1 in
segment 1, F x2 in segment 2, and so on.
With the force now constant in each segment, we can find the work done
within each segment by using Eq. 7-7. Here f ϭ 180Њ, and so cos f ϭ Ϫ1. Then
the work done is ϪF x1 ⌬x in segment 1, ϪF x2 ⌬x in segment 2, and so on. The net
work W s done by the spring, from x i to x f , is the sum of all these works:
(7-22)
where j labels the segments. In the limit as ⌬x goes to zero, Eq. 7-22 becomes
(7-23)
From Eq. 7-21, the force magnitude F x is kx.Thus, substitution leads to
(7-24)
ϭ (Ϫ
1
2 k)[x
2
] x i
x f ϭ (Ϫ
1
2 k)(x f
2 Ϫ x i
2
).
W s ϭ ͵
x f
x i
Ϫkx dx ϭ Ϫk ͵
x f
x i
x dx
W s ϭ ͵
x f
x i
ϪF x dx.
W s ϭ ͚ ϪF xj ⌬x,