3221(9.0 10 N/m)[(0.050 m)( 0.050 m) ]0 J.2sW=×−−=(d) Moving from 5.0 cmix=+to9.0 cmx=−, we have 21(9.0 10 N/m)[(0.050 m)( 0.090 m) ]25 J.2sW−−=−27. From Eq. 7-25, we see that the work done by the spring force is given by 221()2sifWkxx.The fact that 360 N of force must be applied to pull the block to x= + 4.0 cm implies that the spring constant is3360 N90 N/cm9.0 10 N/m4.0 cmk
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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.