hw7_solns

hw7_solns - Answer, Key Homework 7 David McIntyre 45123 Mar...

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Answer, Key – Homework 7 – David McIntyre – 45123 – Mar 25, 2004 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore making your selection. The due time is Central time. Chapters 6 and 7 problems. 001 (part 1 oF 1) 0 points A cheerleader liFts his 70 . 9 kg partner straight up o± the ground a distance oF 0 . 594 m beFore releasing her. The acceleration oF gravity is 9 . 8 m / s 2 . IF he does this 27 times, how much work has he done? Correct answer: 11143 . 5 J. Explanation: The work done in liFting the cheerleader once is W 1 = mg h = (70 . 9 kg)(9 . 8 m / s 2 )(0 . 594 m) = 412 . 723 J . The work required to liFt her n = 27 times is W = nW 1 = (27)(412 . 723 J) = 11143 . 5 J . 002 (part 1 oF 5) 0 points In the parallel spring system, the springs are positioned so that the 44 N weight stretches each spring equally. The spring con- stant For the leFt-hand spring is 3 . 5 N / cm and the spring constant For the righ-hand spring is 5 . 5 N / cm . 3 . 5N / cm 5 44 N How Far down will the 44 N weight stretch the springs? Correct answer: 4 . 88889 cm. Explanation: Let : k 1 = 3 . 5 N / cm , k 2 = 5 . 5 N / cm , and W = 44 N , The springs stretch the same amount x be- cause oF the way they were positioned. Then F 1 = k 1 x and F 2 = k 2 x , so the Force equation For the suspended mass is X F up = X F down k 1 x + k 2 x = W x = W k 1 + k 2 = 44 N 3 . 5 N / cm + 5 . 5 N / cm = 4 . 88889 cm . Dimensional analysis For x : N N / cm = N · cm N = cm 003 (part 2 oF 5) 0 points In this same parallel spring system, what is the e±ective combined spring constant k parallel oF the two springs? Correct answer: 9 N / cm. Explanation: IF the springs were one spring, that spring would react with a Force F = - k x where F = - W due the law oF action and reaction, we have k parallel = - W - x = W x = 44 N 4 . 88889 cm = 9 N / cm , which happens to be the sum oF the individual constants k parallel = k 1 + k 2 = 3 . 5 N / cm + 5 . 5 N / cm = 9 N / cm .
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Answer, Key – Homework 7 – David McIntyre – 45123 – Mar 25, 2004 2 004 (part 3 of 5) 0 points Now consider the same two springs in series. 3 . 5N / cm 5 44 N What distance will the spring of constant 3 . 5 N / cm stretch? Correct answer: 12 . 5714 cm. Explanation: In the series system, the springs stretch a diFerent amount, but each carries the full weight W = 44 N. W = k 1 x 1 and x 1 = W k 1 = 44 N 3 . 5 N / cm = 12 . 5714 cm . 005 (part 4 of 5) 0 points In this same series spring system, what dis- tance will the spring of constant 5 . 5 N / cm stretch? Correct answer: 8 cm.
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hw7_solns - Answer, Key Homework 7 David McIntyre 45123 Mar...

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