2220(17.0 m/s)(9.80 m/s )(42.0 m)26.5 m/s.Bvvgh=+=+=(c) Similarly, 202(17.0 m/s)2(9.80 m/s )(42.0 m)33.4 m/s.Cgh=+=+=(d) To find the “final” height, we set Kf= 0. In this case, we have KU KUmvmghmghfff0002120+=++which yields 2202(17.0 m/ s)42.0 m56.7 m.(9.80m/s)fvhhg=+ =+=(e) It is evident that the above results do not depend on mass. Thus, a different mass for the coaster must lead to the same results. 13. We use Eq. 8-17, representing the conservation of mechanical energy (which neglects friction and other dissipative effects).
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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.