3221(2.0 10 N/m)[(0.050 m)(0.040 m) ]0.90 J.2sW=×−=(b) Moving from 5.0 cmix=+to2.0 cmx=−, we have 21( 0.020 m) ]2.1 J.2sW−−=(c) Moving from ixtox, we have 21( 0.050 m) ]0 J.2sW−−=31. The work done by the spring force is given by Eq. 7-25: 221()2sifWkxx.The spring constant kcan be deduced from Fig. 7-37 which shows the amount of work done to pull the block from 0 to
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.