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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.

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Presentation on theme: "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."— Presentation transcript:

1 Sect. 7-3: Work Done by a Varying Force

2 Work Done by a Varying Force For a particle acted on by a varying force, clearly is not constant! For a small distance  ℓ 1 along the curve, the work done is approximately W 1 = F 1  ℓ 1 cosθ 1 For a small distance  ℓ 2 the work done is approximately W 2 = F 2  ℓ 2 cosθ 2 For a small distance  ℓ i, along the curve, the work done is approximately W i = F i  ℓ i cosθ i The total work over 7 segments is approximately

3 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.

4 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:

5 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 x i to x f ) For those who don’t understand integrals, this is THE AREA under the F vs. x curve

6 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” F s is: F s = -kx (F s > 0, x 0) This is known as Hooke’s “Law” (but it isn’t really a law!)

7 Applied Force F app is equal & opposite to the force F s exerted by block on spring: F s = - F app = -kx

8 Force Exerted by a Spring on a Block Force F s varies with block position x relative to equilibrium at x = 0. F s = -kx spring constant k > 0 x > 0, F s < 0  x = 0, F s = 0  x 0  F s (x) vs. x 

9 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 2 nd Law says: ∑F y = 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

10 W = (½)kx 2  Relaxed Spring Spring constant k x = 0   x  W W In (a), the work to compress the spring a distance x: W = (½)kx 2 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

11 Plot of F vs. x. The work done by the person is equal to the shaded area.

12 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?

13 Example 7-6: Force as a function of x where F 0 = 2.0 N, x 0 = 0.0070 m, and x is the position of the end of the arm. If the arm moves from x 1 = 0.010 m to x 2 = 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


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