The information on this diagram is to be used for the accompanying questions. Using the information shown on the diagram on the other page, determine the correct answers to the following questions. 1 How much power was used to pull the coaster car up the first hill? (in watts and horsepower.) 2 How fast is the car traveling at location A if it coasted down the back side of the first hill? 3 How many g's does the rider feel as he enters the bottom of the next element if it's radius is 60.0 m, at location B? 4 How fast is the car traveling when it reaches the banked curve? 5 If the banked curve is at a 52.3° angle with the horizontal, then what is the radius of the banked curve? 6 How many g's does the rider feel in the banked curve? 7 How many g's does a rider feel as he enters the bottom of the loop? 8 How fast is the rider traveling at the top of the loop? 9 How many g's does the rider feel at the top of the loop? 10 As the car enters the station at the end of the ride, its brakes are applied. How much power do the brakes exert on the car? (in watts and horsepower.) 1 Use Energy relationships and power to solve vo = 5.50 m/s vf = 4.44 m/s h = 62.5 m t = 45.8 s m =6500 kg DET = [KEf + PEf] - [KEo + PEo] DET = [(1/2)m(vf)2 + mg(hf)] - [(1/2)m(vo)2 + 0] DET = [(1/2)(6500)(4.44)2 + (6500)(9.80)(62.5)] - [(1/2)(6500)(5.50)2] DET = 64069.2 + 3981250 - 98312.5 DET = 3947006.7 J Power = 86,179.186 watts ...746 w = 1 hp Power = 115.52 hp 2 Use Energy relationships to solve vo = 4.44 m/s ...at the hill top vf = ???? ...at the hill bottom h = 62.5 + 22.1 = 84.6 m ET @ the hill top = ET @ the hill bottom KE + PE = KE + PE (1/2)mv2 + mgh = (1/2)mv2 + 0 (1/2)v2 + gh = (1/2)v2 + 0 (1/2)(4.44)2 + (9.80)(84.6) = (1/2)v2 838.94 = (1/2)v2 v = 40.96 m/s 3 Use circular motion relationships to solve v = 45.5 m/s, r = 60.0 m g's felt at the bottom = ac[in gs] + 1 g g's felt at the bottom = 3.52 + 1 g g's felt at the bottom = 4.52 g's 4 Use Energy relationships to solve ET @ "B" = ET @ "A" KE + PE = KE + PE (1/2)mv2 + 0 = (1/2)mv2 + mgh (1/2)v2 = (1/2)v2 + gh (1/2)(45.5)2 = (1/2)v2+ (9.80)(60.0) 447.125 = (1/2)v2 v = 29.904 m/s 5 Use banked curves v = 29.90401311 m/s ...from #4 q = 52.3° 6 Use banked curves g's felt = 1.26 7 Use circular motion relationships to solve v = 45.5 m/s ...because it is at the same height as location "B." r = 15 m g's felt at the bottom = ac[in gs] + 1 g g's felt at the bottom = 14.083 + 1 g g's felt at the bottom = 15.0 g's 8 Use Energy relationships to solve ET @ bottom = ET @ top KE + PE = KE + PE (1/2)mv2 + 0 = (1/2)mv2 + mgh (1/2)v2 = (1/2)v2 + gh (1/2)(45.5)2 = (1/2)v2+ (9.80)(30.0) 741.125 = (1/2)v2 v = 38.5 m/s 9 Use circular motion relationships to solve v = 38.5 m/s ...from #8 r = 15 m g's felt at the top = ac[in g's] - 1 g g's felt at the top = 10.083 - 1 g g's felt at the top = 9.08 g's 10 Use energy relationships and power to solve vo = 25.05 m/s vf = 3.03 m/s d = 10 m t = 45.8 s m =6500 kg DET = [KEf + PEf] - [KEo + PEo] DET = [(1/2)m(vf)2 + 0] - [(1/2)m(vo)2 + 0] DET = (1/2)(6500)(25.05)2 - (1/2)(6500)(5.50)2 DET = 2010103.875J t = 0.712250712 s Power = 2822185.841watts ...746 w = 1 hp Power = 3783.09 hp   If you use or find this page useful or have any comments, please contact the author so maybe he'll do more. Author: Tony Wayne "ROLLER COASTER PHYSICS" TABLE OF CONTENTS ... PHYSICS PAVILION TABLE OF CONTENTS <--PREVIOUS SECTION ... NEXT SECTION --> A special thanks to VASTfor hosting our web site.