Getting the Coaster Started Something has to be done to get the coaster started. In our previous example energy, power, has to added get the coaster up to 8.8 m/s. This is done by doing work on the coaster. A simplified definition of work would be force times displacement when the force and displacement go in the same direction. [This chapter will not go into all the details of calculating work.] Suffice it to say that when the force acting on the coaster and the displacement of the coaster are in the same direction, work adds energy to the coaster. When the force acting on the coaster and the displacement of the coaster are in opposite directions, work removes energy from the coaster. Work = (Force)(Displacement) W = Fd Where "work" is measured in joules, J. "Force" is measured in Newtons, N, and "displacement" is measured in meters, m. Example 1 What is the velocity of the train after being catapulted into motion? Solution: ET(BEGINNING) + Work = ET(TOP OF 1st HILL) KE + PE +W = KE +PE (1/2)mv2 + mgh + Fd = (1/2)mv2 + mgh (1/2)3000(0)2 + 3000(9.8)(3000) + 8800(12.5)= (1/2)3000v2 + 3000(9.8)(0) Substitute the numbers at each location 110,000 = (1/2)(3000)v2 73.333 = v2 v = 8.6 m/s ... at the end of the catapult. As an aside you can calculate the acceleration of the rider from kinematics equations. (For the curious the acceleration is 7.0 m/s2) Example 2 What is the velocity of the train after being catapulted into motion? ` Solution: ET(BEGINNING) + Work = ET(TOP OF 1st HILL) KE + PE +W = KE +PE (1/2)mv2 + mgh + Fd = (1/2)mv2 + mgh (1/2)3000(5.2)2 + 3000(9.8)(0) + 19,715.464(145.5) = (1/2)3000v2 + 3000(9.8)(95) 40560 + 2868600.012 = (1/2)(3000)v2 + 2793000 116160.012 = (1/2)(3000)v2 77.44 = v2 v = 8.8 m/s ... at the end of the catapult. Example 3 A roller coaster train of mass 3.0 X 103 kg rolls over a 11.5 m high hill at 8.34 m/s before rolling down into the station. Once in the station, brakes are applied to the train to slow it down to 1.00 m/s in 5.44 m. (a) What braking force slowed the train down? (b) How much time did it take to slow the train down? (c) What was the acceleration of the train in g's? Solution: (a) ET(HILL) = ET(@ 1 m/s) + Work KE + PE = KE +PE +W (1/2)mv2 + mgh = (1/2)mv2 + mgh + Fd (1/2)3000(8.34)2 + 3000(9.8)(11.5) = (1/2)3000(1)2 + 3000(9.8)(0) + F(5.44) Substitute the numbers at each location 442433.400 = 1500 + 5.44F F = 81053.9 N ...force to slow down the train (b) Calculate the velocity as the train enters the station. Use this velocity to calculate the time. ET(HILL) = ET(@ STATION ENTRANCE) No work is done because no force acts between the two locations KE + PE = KE +PE (1/2)mv2 + mgh = (1/2)mv2 + mgh (1/2)3000(8.34)2 + 3000(9.8)(11.5) = (1/2)3000v2 + 3000(9.8)(0) Substitute the numbers at each location 442433.400 = 1500v2 v = 17.174 v = 17.2 m/s ...as the train enters the station. The time is calculated from vo = 17.174 m/s vf = 1.00 m/s x = 5.44 m t = ? t = 0.599 sec (c) Calculate the acceleration in m/s2. Then convert it into g's. F= ma 81053.934 N = (3000 kg)a a = 37.018 m/s2 a = 37.018 m/s2/ 9.80 m/s2 = 3.78 g's ... Yeow! That's a big jerk on the passengers into the restraining harness. 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.