• O • Is University of South Alabama XY WileyPLUS Ahmad * G Chegg Study Guided S
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• O • Is University of South Alabama XY WileyPLUS Ahmad * G Chegg Study Guided Solution X D Travis Scott - Butterfly Ette 4 x + ca Secure | https//edugen.wileyplus.com/edugen/student/mainfruni Y * NEW 1 I Apps R WileyPLUs a Differential calculus LS University of South. Apple C iCloud Yahoo b Bing w Wikipedia F Facebook , y Twitter in LinkedIn D The Weather Channel O Yelp >> other Bookmarks wleyPLUS: MywlePLUSI Help I contact Us I Log Out WileyPLUS haiday, Fundamentals of Physics10e Calculus-based Physics I & II (PH 201-202) Home Read, Study & Practice Assignment Gradebook ORION Downloadable eTextbook Assignment > Open Assignment PULL SCREEN PRINTER VERSION 1 BACK NTS ASSIGNMENT RESOURCES 201-6 Chapter 05, Problem o02 Ch. 056 Two horizontal forces act on a 8.8 kg chopping block that can slide over a frictionless kitchen counter, which lies in any plane. One force is F1 = (7.9 N 8.6 N. the ) + (JFind acceleration Chapter 04. Problem 05 .of the chopping block in unit-vector notation when the other force is (a) F2 - -79m p-- as N D. (b) F2= (- ramp+ (asonD, and to ) Ch. 006 F2 = (7.9 N / +- 8.6 N | Review Score Review Results by Study objective (. (a) Number Units (b) Number Units (c) Number Units licensa Agreement privacy Policy ID 2000-201Zohn Wiley & Sons, Inc. All Rights Reserved. A Division of ohn Wiley & Sons, Inc. Version 4.24.1.20 https://edugen-wileyplus.com/ediugen/shared/assignment/test/agist uni?id-asnmt 204900Explanation / Answer
given, mass of object m = 8.8 kg
the motion is only in xy plane
F1 = 7.9 i + 8.6 j
a. F2 = -7.9 i - 8.6 j
so Net force, F = F1 + F2 = 7.9 i + 8.6 j -7.9 i - 8.6 j = 0 N
let the acceleration be a
then from newton's second law
ma = F = 0
a = 0 m/s/s
b. F2 = -7.9 i + 8.6 j
so Net force, F = F1 + F2 = 7.9 i + 8.6 j -7.9 i + 8.6 j = 17.2 j
let the acceleration be a
then from newton's second law
ma = F = 17.2 j
a = 17.2 j/8.8 = 1.954 m/s/s [ along +y axis]
c. F2 = 7.9 i - 8.6 j
so Net force, F = F1 + F2 = 7.9 i + 8.6 j + 7.9 i - 8.6 j = 15.8 i N
let the acceleration be a
then from newton's second law
ma = F = 15.8 i
a = 15.8 i / 8.8 = 1.795 m/s/s [ along +ve x axis]
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