Sect. 7-3: Work Done by a Varying Force. Work Done by a Varying Force For a particle acted on by a...
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Transcript of Sect. 7-3: Work Done by a Varying Force. Work Done by a Varying Force For a particle acted on by a...
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Sect. 7-3: Work Done by a Varying Force
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Work Done by a Varying ForceFor a particle acted on by a varying force, clearly
is not constant!
For a small distance ℓ1 along the curve, the work
done is approximately W1 = F1 ℓ1 cosθ1
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
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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.
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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:
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• 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
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Work Done by an Ideal Spring ForceAn 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!)
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Applied Force Fapp is equal & opposite to the forceFs exerted by block on spring: Fs = - Fapp = -kx
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Force Exerted by a Spring on a Block
Force Fs varies with block position x relative to equilibrium at x = 0. Fs = -kx
spring constant k > 0
x > 0, Fs < 0
x = 0, Fs = 0
x < 0, Fs > 0
Fs(x) vs. x
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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 = 0so, 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
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W = (½)kx2
Relaxed Spring Spring constant k
x = 0
x
W
W
In (a), the work to compress the spring a distance x: W = (½)kx2
So, the spring stores potential energy in this amount.
W In (b), the spring does work on
the ball, converting it’s stored potential energy into
kinetic energy.
W
W
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Plot of F vs. x. The work done by the person is equal to the shaded area.
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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?
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Example 7-6: Force as a function of x
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?
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