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Quiz:  Run Multiple Choice Generator for 'course' 201_core, using choice #2.

General College Physics 6th Edition Asst:

Chapter 4 problems 3, 4, 5, 6, 8, 12, 16, 19, 20, 21, 30, 32, 37, 47, 49, 51, 56

University Physics:  Text Problems as Assigned in thru Asst 14 on Homepage.

Experiment:

Assuming that your rubber band has a force vs. slope graph with slope .8 N / cm determine the area beneath the force vs. position graph from x = 0 to x = .5 cm, from x = .5 cm to x = 1 cm, from x = 1 cm to x = 1.5 cm and from x = 1.5 cm to x = 2 cm.  Completely label a trapezoidal graph indicating all relevant quantities, including accumulated areas..

Project the half-meter stick across the table using pullbacks of 1 cm, 2 cm, 3 cm and 4 cm.  See how far the stick slides in each case.

Using the same pullbacks as before, project the stick off the edge of the table, with the center of the stick at the edge of the table when you begin your pullback.  Determine as best you can how far the center of the stick falls from the time it leaves the edge of the table until it hits the floor.

Measure the dimensions of the half-meter stick.  Assuming that the density of the stick is .6 grams / cm^3 find its mass and its weight.

From the projectile behavior of the stick figure out how much KE the stick attained as a result of the action of the rubber band.

Assuming that the stick attained the same KE for each pullback on the tabletop figure out the average force exerted by friction, and express the average force as a percent of the weight of the stick.

See the discussion below.

Univ. Phy: 

• Assuming force function F(x) = .8 N / m * x, where x is the amount of stretch in the rubber band, use integration to find the work done between 0 cm and .5 cm, between .5 cm and 1 cm, between 1 cm and 1.5 cm and between 1.5 cm and 2 cm.  Compare with the results of your trapezoidal approximations.
• Integrate F(z) with respect to z from z = 0 to z = x for each of the following values of x:  .5 cm, 1 cm, 1.5 cm and 2 cm.  What numbers on your trapezoidal graph do your results correspond to?

Discussion of the F vs. x graph:

A graph of rubber band force vs. stretch is approximated by a function with slope .8 N / cm, passing through the origin (since 0 stretch implies 0 tension the graph will go through the origin).

To approximate the area between x = 0 cm and x = 1.5 cm we proceed as follows:

• The run from x = 0 cm to x = 1.5 cm is 1.5 cm and the slope of the graph is 1.5 N / cm.
• The rise between two points is rise = run * slope. 
• So between x = 0 and x = 1.5 cm the rise is 1.5 cm * .8 N / cm = 1.2 N * cm / cm = 1.2 N.
• Starting from the origin and moving through run 1.5 cm and rise 1.2 N we arrive at the point (.5 cm, 1.2 N).
• We construct a trapezoid (which is actually a triangle in this case) from the x = 0 point to the x = 1.5 cm point.
• We label the right-hand altitude 1.2 N, since that is the value of the vertical coordinate (the force is measured in the vertical direction).  We also label the x axis with 0 at the left-hand extreme of the trapezoid and 1.5 at the right.
• The area beneath the trapezoid is the product of its average 'altitude' and its width.
• Trapezoid altitudes are 0 N and 1.2 N, so its average altitude is (0 N + 1.2 N) / 2 = .6 N.
• Trapezoid width is from x = 0 cm to x = 1.5 cm, a width of 1.5 cm.
• Therefore trapezoid area is ave altitude * width = .6 N * 1.5 cm = .9 N cm.

University Physics Students:

• To get the area beneath the F(x) vs. x curve you integrate F(x) from the left-hand x value  to the right-hand x value.
• The F(x) function given below has vertical intercept 0 and slope .8 N / cm, so F(x) = m x + b = .8 N / cm + 0, or just F(x) = .8 N / cm.
• For example if we want the area beneath the curve from x = .5 cm to x = 1.5 cm  we do

area = integral( .8 N / cm * x, with respect to x, x from .5 to 1.5 cm.)

• .8 N / cm is a constant so we have the .8 N / cm * the integral of x from x = .5 cm to x = 1.5 cm.  An antiderivative of x is x^2 / 2, so the integral of x is just the change in the value of x^2 / 2.  The change in this expression is (1.5 cm)^2 / 2 - (.5 cm)^2 / 2 = 2.0 cm^2 / 2 = 1.0 cm^2.  If we multiply this by the .8 N / cm we get .8 N / cm * (1 cm^2) = .8 N * cm.
• This result matches the area under the graph from x = .5 cm to x = 1.5 cm, as you can easily verify.
• Be sure you have reviewed integration of power functions, polynomials and sine and cosine functions, as well as the Chain Rule.