470 Oscillatory Motion*P15.74(a)The block moves with the board in what we take as the positive xdirection, stretching thespring until the spring force −kxis equal in magnitude to the maximum force of staticfriction µµssnmg=. This occurs at xmgks=µ.(b)Since vis small, the block is nearly at the rest at this break point. It starts almost immediatelyto move back to the left, the forces on it being −kxand +kmg. While it is sliding the netforce exerted on it can be written as−+==− −FHGIKJ=−kxmgkxkmgkkxmgkkxkkkrelwhere xrelis the excursion of the block away from the point kmgk.Conclusion: the block goes into simple harmonic motion centered about the equilibriumposition where the spring is stretched by kmgk.(d)The amplitude of its motion is its original displacement, Amgkmgksk. It first comes torest at spring extension kksmgkAmgk−=−2bg. Almost immediately at this point it
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .