15_R_7 - 7. r = a sin ti + a cos tj + btk, (0 t 6 ) r(0) =...

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Unformatted text preview: 7. r = a sin ti + a cos tj + btk, (0 t 6 ) r(0) = aj, r(6 ) = aj + 6 bk. a) The force F = -mgk = - (mgz) is conservative, so the work done by F as the bead moves from r(6 ) to r(0) is t=0 z=0 W = t=6 F dr = -mgz z=6 b = 6 mgb. b) v = a cos ti-a sin tj+bk, |v| = a 2 + b2 . A force of constant magnitude R opposing the motion of the bead is in the direction of -v, so it is F = -R v R = - v. |v| a 2 + b2 Since dr = v dt, the work done against the resistive force is 6 W = 0 R a2 + b2 |v|2 dt = 6 R a 2 + b2 . ...
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This note was uploaded on 03/08/2012 for the course MATH 120 taught by Professor Onurfen during the Spring '12 term at Middle East Technical University.

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