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i Chrome File Edit View History Bookmarks People Window Help x-e Chegg Study I Guided Solutio x y C secure https://edugen.wileyplus.com/edugen/student/mainfr.uni 46 15% D' us. Mon 9:11 PM Q NNetfix WileyPLUS Ahmad Apps V. WileyPLUS Differential calculus . LA University of South D Apple Cloud Yahoo Bing W Wikipedia Facebook y Twitter LinkedIn Other Bookmarks WileyPLUS Halliday, Fundamentals of Phyaics, 10 Calculus-based Physics I & II (PH 201-202) Home Read,Study & Practice Gradebook ORION Downloadable eTextbook Assianment> Open Assignment k of ca.pdf FULL SCREENPRINTER VERSTON BACK NEXT ASSIGNMENT RESOURCES Chapter 07, Problem 016 201-10 A 7.3 kg object is moving in the positive x direction, when it passes through x = 0, a constant force directed along the axis begins to act on it. The figure here gives its kinetic energy K versus position x as it moves rrom x = 0 to x 5.0 m; K0 = 29 . The force continues to act. what is speed v when the object moves back through X =-3.1 m? 040 K (J) Ko 016 cpp Review Score Review Results by Study Unit smnto the tolerance is +/-2% GO TUTORIAL LINK TO TEXT LINK TO SAMPLE PROBLEMVIDEO MINI-LECTURE n Shot 47.15 Question Attempts: 0 of 3 used SAVE FOR LATER SUBMIT ANSWER shot j All Rights Reserved. A Division of Version 4.24.1.23Explanation / Answer
from work enery theorem
work done = F*dx = change in KE
F*(5-0) = 0 - 29
F = -5.8 N
from x1 = 5 m to x2 = -3.1 m
Work = change in kinetic energy
F*dx = (1/2)*m*(v2^2 - v1^2)
-5.8*(-3.1-5) = (1/2)*7.3*(v2^2-0)
v2 = 3.58 m/s ,<<<<<------ANSWER
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