Welcome to the website of ME 274 for the Fall 2008 semester. On this site you can view blog posts, add your own blog posts and add comments to existing posts. In addition to the blog are links to course material:
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6 comments:
I am having problems coming up with the right energy lost. how do you equate the kinetic energies? the only equation I could come up with is (.5m(a+b)v(f)^2)-(.5m(a)v(a)^2+.5m(b)v(b)^2)=delta(E). Is that the correct equation but I am messing up the conversions between mi/hr and ft-lb or did I mess up somewhere else?
remember that lbs in this book refer to weight, not mass. You also have to convert speed to ft/s for the kinetic energy equation to work out to ft-lb.
hmm...I converted all units to ft/s and lb mass but I am still get an answer that is 192 lbs off. The equation I am using is:
delta E = (0.5(mA + mB)vf^2) - [ (0.5(mA)(vA)^2) + (0.5(mB)(vB)^2) ]
any ideas what I am doing wrong.
sri-
I had the exact same problem. It's just caused by rounding the vf term before plugging it into your equation.
I agree with Panda Watch in that the problem likely lies in rounding. When I carry through with the accuracy of my calculator and using g = 32.2 ft/sec^2, I agree with the solution in the book.
Since your answer is off by 192 ft-lb, your answer is accurate to only 1 significant digit. This means that at least some of your calculations were made using at most 2 significant digits.
Look back on the rounding that you used in using v_2 or in your number of digits in g.
ah, right! Thanks!
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