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PHY 241
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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My question is regarding the test. I was reviewing possible questions and did not understand the notation for a particular problem. The problem is in the next box.
My question is about k= 190 N/m^.65. What is raised to the .65? Is k = 190^.65 N/m or 190^(-.65)N/m or something else. From past experiance I might assume k=190N/m while x^.65 meters. hence the function [replacing 190 for k] is F(x) = 190 * x^(.65) which is in units of N/m * m
By the order of operations, only the meter unit is raised to the .65.
If x is in meters, then x^.65 will be in units of meters^.65. To get a result in Newtons, you're going to need k to be in N / m^.65. For example if x = 8 m, then 190 N / m^.65 * (8 m)^.65 = 190 N / m^.65 * 2.7 m^.65 = 540 N, or so.
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Problem Number 3
The force exerted by a rubber band at stretch x is given by the function F(x) = k x^ .65, with k = 190 N / m^ .65.
If an object is accelerated from rest by the rubber band which is initially stretched by .185 meters, and as a result slides across a horizontal surface against which it must exert a force of 29 Newtons, then how far will the object slide (assume that none of the potential energy in the rubber band is dissipated and that the mass of the rubber band is negligible)?
On this one you would integrate k x^.65 from 0 m to .185 m, integrating force vs. stretch, in order to get the PE of the rubber band.
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I hope this makes sense I just need to be sure this is a correct understanding of the question.
Check the notes and let me know if you have more questions.
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