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  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

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 x i to x f ) 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” 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!)

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

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 

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

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

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 where F 0 = 2.0 N, x 0 = m, and x is the position of the end of the arm. If the arm moves from x 1 = m to x 2 = 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