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Sect. 7-3: Work Done by a Varying Force
Work Done by a Varying Force
For a particle acted on by a varying force, clearly
is not constant!
For a small distance ℓ2 the
work done is approximately
W2 = F2 ℓ2 cosθ2
For a small distance ℓi,
along the curve, the work
done is approximately
Wi = Fi ℓi cosθi
The total work over 7
segments is approximately
For a small distance ℓ1 along the curve, the work
done is approximately W1 = F1 ℓ1 cosθ1
For a force that varies, the work can be approximated by
dividing the distance up into small pieces, finding the
work done during each, and adding them up.
In the limit that the pieces become infinitesimally
narrow, the work is the area under the curve, which is
the integral of Fcosθ over the distance ℓ
Or:
• See text for details. Requires that you know simple integral calculus.
• In one dimension, for F = F(x), the bottom line is that the work
done is the integral of the F vs. x curve:
W = ∫ F(x) dx
(limits xi to xf)
• For those who don’t understand
integrals, this is THE AREA
under the F vs. x curve
Work Done by an Ideal Spring Force
An ideal spring is characterized by a spring
constant k, which is measure of how “stiff”
the spring is. The “restoring force” Fs is:
Fs = -kx
(Fs > 0, x < 0; Fs < 0, x > 0)
This is known as Hooke’s “Law”
(but it isn’t really a law!)
Applied Force Fapp is equal & opposite to the force
Fs exerted by block on spring: Fs = - Fapp = -kx
Force Exerted by a Spring on a Block
Force Fs varies with
block position x
x > 0, Fs < 0
relative to equilibrium
at x = 0. Fs = -kx

x = 0, Fs = 0 
spring constant k > 0
x < 0, Fs > 0 
Fs(x) vs. x 
Example: Measuring k for a Spring
Hang a spring vertically. Attach
an object of mass m to the lower
end. The spring stretches a
distance d. At equilibrium,
Newton’s 2nd Law says:
∑Fy = 0
so, mg – kd = 0 or mg = kd
If we know m, & measure d,
 k = (mg/d)
Example:
d = 2.0 cm = 0.02 m
m = 0.55 kg
 k = 270 N/m
x=0

 Relaxed Spring
Spring constant k

W=
(½)kx2
x
In (a), the work to compress the
2
 spring a distance x: W = (½)kx
So, the spring stores potential
energy in this amount.
W
W
W In (b), the spring does work on
the ball, converting it’s stored
potential energy into
kinetic energy.
W
W
Plot of F vs. x. The work done by the person
is equal to the shaded area.
Example 7-5: Work done on a spring
a. A person pulls on a spring, stretching it x = 3.0 cm,
which requires a maximum force F = 75 N. How
much work does the person do?
b. Now, the person compresses the spring x = 3.0 cm,
how much work does the person do?
Example 7-6: Force as a function of x
A robot arm that controls the position of a video camera in an
automated surveillance system is manipulated by a motor that
exerts a force on the arm. The dependence of the force on the
position x of the robot arm is measured & found given by
where F0 = 2.0 N, x0 = 0.0070 m, and x is the position of the
end of the arm. If the arm moves from x1 = 0.010 m to x2 =
0.050 m, how much work did the motor do?