SCI103 Phase 5 Lab Report: Exploring Potential and Kinetic Energy

A lab report in Physics 103, exploring potential and kinetic energy concepts.

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SCI103 Phase 5 Lab Report: Exploring Potential and Kinetic EnergyName:Michelle KleszynskiDate:2/3/13Instructor’s Name:NemmersAssignment:SCI103 Phase 5 LabReportTITLE: Potential and Kinetic EnergyINSTRUCTIONS: Enter the Virtual Lab and conduct the experiments provided.Pleasetype your answers. When your lab report is complete,submit it to the SubmittedAssignments area of the Virtual Classroom.PartIAnswer the following questions while in thePhase 5lab environment.Section 1From the left of the screen to the right, thered ballshave acenter of mass placed at20 feet, 15 feet, and 10 feethighrespectively.1.Suppose each red ball weighs 20lbs. Find the potential energy(PE)foreach balloneach ramp.PE=mgh + 20lbs x 32.17x 20 ftPE=12868 foot-PoundsPE=mgh 20 lbs x 32.17 x 15 ftPE= 9651 foot-poundsPE= mgh 20 lbs x 32.17 x 10 ftPE= 6434 foot pounds2.Predict the maximum speed of the ball on each ramp. How would this speedchange if each ball’s mass was doubled?KE=1/2mv2KE= ½20lbsx 12868 squared= ½ would be 10/12868= 1286.8 squared=KE=35.87 m/secKE=1/2 20lbs x 9651 squared= ½ would be 10/9651=965.1 squared=KE=31.07m/secKE= ½ 20lbs x 6434 squared= ½ would be 10/6434= 634.3 squaredKE= 25.37 m/secIf the mass what doubled the KE would = 20/12868 KE=25.37m/sec20/9651 KE= 22 m/sec20/6434 KE= 17.93 m/secThe speed would go down when you add more weight.3.On ramp 1, calculate the balls potential and kinetic energy at 16 feet.PE=mgh 20lbs x 32.17 x 16ft PE= 10294.4 foot-poundsKE= ½ 20lbs x 10294.4 squared+ ½ wouldbe 10/10294.4 =1029.44 squaredKE=32.08 m/sec

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