dont copy solve it properly with all steps A force F = 10 sin at N acts on a displacement of x = 2 sin (nt – n16)m. Determine (a) the work done during the first 6 s; (b) thework done during the first 12s.Solution: Rewriting Eq. (3.7-1) as W=[Fx°dt and substituting F = Fo sint and x = X sin (t- ) gives the work done per cycle ofW=nF0X0 sin ởFor the force and displacement given in this problem, Fo = 10 N, X = 2 m,= n/6and the period= 2s. Thus, in the 6 s specified in (a),three complete cycles take place, and the work done isW=3(nF0X0sin o)=3rx10x2 × sin 30°=94.2 N •mThe work done in part (b) is determined by integrating the expression for work between the limits 0 and 12s.W=WF0X0[cos 30°f01/2sin nt cos at di+sin 30°f01/2sin nt dt]=Tx10x2[-0.8664rcos 2rt+0.50(12–sin 2rt47)]01/2=16.51N:1

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Answered: A force F = 10 sin at N acts on a…

A force F = 10 sin at N acts on a displacement of x = 2 sin (t – n16)m. Determine (a) the work done during the first 6 s; (b) the work done during the first 12s.

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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