SOL17 - Thevalingam Donald – Homework 17 – Due midnight...

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Unformatted text preview: Thevalingam, Donald – Homework 17 – Due: Jan 29 2007, midnight – Inst: Eslami 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points It takes 4 . 19 J of work to stretch a Hooke’s-law spring 13 cm from its unstressed length. How much the extra work is required to stretch it an additional 7 . 91 cm? Correct answer: 6 . 65015 J. Explanation: We begin by determining the value of k from the energy equation solved for k , k = 2 U x 2 = 2(4 . 19 J) (0 . 13 m) 2 = 495 . 858 J / m 2 We can now determine the energy at the ad- ditional displacement, U add = 1 2 k x 2 add = 1 2 495 . 858 J / m 2 (0 . 13 m + 0 . 0791 m) 2 = 10 . 8402 J The extra work required is just the difference in energy between the two displacements. W = Δ U = U add- U = 10 . 8402 J- 4 . 19 J = 6 . 65015 J keywords: 002 (part 1 of 2) 10 points The force required to stretch a Hooke’s-law spring varies from 0 N to 78 . 3 N as we stretch the spring by moving one end 13 . 9 cm from its unstressed position. Find the force constant of the spring. Correct answer: 563 . 309 N / m. Explanation: At the extended position, we can write Hooke’s law, ( F = k x ) solved for k . k = F x = 78 . 3 N . 139 m = 563 . 309 N / m 003 (part 2 of 2) 10 points Find the work done in stretching the spring. Correct answer: 5 . 44185 J. Explanation: Having determined k , we can determine the work done in stretching the spring. This is just the energy stored in the spring, W = 1 2 k x 2 = 1 2 (5 . 44185 J) (0 . 139 m) 2 = 5 . 44185 J keywords: 004 (part 1 of 1) 10 points...
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This note was uploaded on 03/04/2010 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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SOL17 - Thevalingam Donald – Homework 17 – Due midnight...

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