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A force F = 10 sin at N acts on a displacement of x = 2 sin (tư – a16)m. Determine (a) the work done during the first 6 s; (b) the work done during the first 12s.

A force F = 10 sin at N acts on a displacement of x = 2 sin (tư – a16)m. Determine (a) the work done during the first 6 s; (b) the work done during the first 12s.

Elements Of Electromagnetics
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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plz do this problem unable to understand explain all steps related to vibration dolve integration also properly and also find work done during 8 seconds
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) the work done during the first 12s. Solution: Rewriting Eq. (3.7-1) as W=[Fx°dt and substituting F = Fo sin t and x = X sin ( t- ) gives the work done per cycle of W=nF0X0 sin ở For 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 is W=3(nF0X0sin o)=3rx10x2 × sin 30°=94.2 N •m The 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
Transcribed Image Text: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) the work done during the first 12s. Solution: Rewriting Eq. (3.7-1) as W=[Fx°dt and substituting F = Fo sin t and x = X sin ( t- ) gives the work done per cycle of W=nF0X0 sin ở For 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 is W=3(nF0X0sin o)=3rx10x2 × sin 30°=94.2 N •m The 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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