CS Chegg CPhy Wiley PLUS Chegg. X C edugen.wileyplus.com /edugen/student/mainfr.
ID: 1356397 • Letter: C
Question
CS Chegg CPhy Wiley PLUS Chegg. X C edugen.wileyplus.com /edugen/student/mainfr.un Home Read, Study & Practice Gradebook ORION Assignment Open Assignment FULL SCREEN PRINTER VERSION BACK ONEXTD ASSIGNMENT RESOURCES Chapter 07, Problem 033 Homework 7 Fall The block in the figure lies on a horizontal frictionless surface, and the spring constant is 46 N/m. Initially, the spring is at its relaxed length and the block is stationary at position x 0 2015 Then an applied force with a constant magnitude of 5.0 N pulls the block in the positive direction of the x axis, stretching the spring until the block stops. When that stopping point is Chapter 07, Concept reached, what are (a) the position of the block, (b) the work that has been done on the block by the applied force, and (c) the work that has been done on the block by the spring 04 force? During the block's displacement, what are (d) the block's position when its kinetic energy is maximum and (e) the value of that maximum kinetic energy Chapter 07, Concept 05 Chapter 07, Concept Block 07 attached V Chapter 07 Problem to spring 001 Chapter 07, Prob 008 Ch ter 07, Prob V Chapter 07. Problem 023 (a) Number Units Ch ter 07, Prob 024 (b) Number 07, Prob Units 031 07, Prob (c) Number Units 033 07, Prob (d) Number Units 036 07, Prob (e) Number Units 038 07, Prob 045 07, Prob 050 Chapter 07, Problem Question Attempts: 0 of 5 used SAVE FOR LATER SUBMIT ANSWER License Agreement l Privacy Policy l 2000-2015 John Wiley & Sons Inc. All Rights Reserved. A Division of John Wiley & Sons Inc Version 4.16.1.7 t/test/aglist uni N100BB 3:50 PM Ask me anything 0/10/2015Explanation / Answer
k = 46 N/m
F = 5 N
a) at equilibrium position : F = kx^2
5 = 46 x^2
x = 0.3297 m
b) work done = force x displacement = F x = 5 * 0.3297 = 1.6485 J
c) work done by spring = - work done by force = - 1.6485 J
d) max kinetic energy is when all of potential energy is converted into kinetic energy ie at position x = 0
e) max kinetic energy = 0.5 k(x)^2 = 0.5 * 46 *0.3297^2 = 2.5 J