Question

he goal of this problem is to engineer a roller-coaster (see Figure 1). Imagine you work, for an amusement park and your boss asks you to buy a spring that could launch a 2,000kg roller-coaster train on the track. In this problem we consider the acceleration of gravity to be g -9.8m/s. art 1: Frictionless track. We first consider a frictionless track. The train is first compressing a spring of constant k and is suddenly released. The compression A. 4m. Calculate the total energy of the train as a function of the spring! constant k. (b) G Calculate the total energy of the train at the top of the first hill (point in Figure 1) as a function of the spring constant and of the velocity ve at this point. What is the minimum spring constant kmin for the train to go over the first hill (point C in Figure 1)? Now, we want the speed at point to be at least 2m/s. What is the minimum value of the spring constant? 8) Assuming the speed of the train at point C is ve = 2m/s, calculate the velocity of the train at point D. .) Part 2: Friction on the track. We now consider that the train experiences kinetic friction (n = 0.2) between point A and point B (10m distant), and between point E and point F (we denote L the distance between E and F). We assume the spring constant is k = 80,000J/m2 and the compression of the spring is Az = 4m. What is the velocity of the train just after being released from the spring (without touching it)? Calculate the kinetic friction force the train experiences between A and 2. (6) w much kinetic energy does the train looses between A and B? Calculate the speed of the train at point B. Joes the train have enough energy to go through the two hills? Explain What is the speed of the train at point E (find the answer without doing any calculations). Explain. The train is now stopping from point E to point F. How far does the train have to travel from point E to be able to stop at point F? We assume the coefficient of kinetic friction to be i = 0.8 when the brakes are pushed. Figure 1: Roller-coaster. lom

Answer 1

Peru D as Total energy = 2K@x)" b) At top of the hill, P. E = mgha 20008 9. 8 % 30 588 00 5. KE- CK Carga) - PE Total energy will be conserved co tx @ng? » mgh - 2x 2000 x 9.8 x 30 nr 2 mgh Como (4) 2 N 73500 sl. unit d) tacon²= P.E.TKE = mght Imp? = 588000 + $ x 2000 x 4 588000 + 4000 - 592 000 74ooo S. I unit. K- 592000X2 = = 16