Question

A car of mass 500. kg on a roller coaster ride is raised from the ground (horizontal dotted line) to point A as shown in the diagram above. When the car passes point A, its speed is 5.0 m/s, and then it rolls freely through the rest of the ride with negligible friction, passing points B, C, D and E. Note that point D is the highest portion of a circular loop, and point E is the lowest portion of that loop. The height of each point (except B and E) is marked as shown (neglect rotational kinetic energy of the car wheels).

(a) Calculate the total (mechanical) energy of the car at point B.

(b) Calculate the speed of the car when it is at point C.

(c) Determine the total (mechanical) energy of the car at the top of the loop (point D)

(d) Calculate the magnitude of the normal force on the car when it passes point D.

(e) Calculate the car’s kinetic energy at point E.

(f) Suppose that, after passing point E, the car reaches a portion of the track which is parallel to the ground and eventually slows down to a stop due to frictional forces. Determine the amount of work done by friction in stopping the car.

500.kg B 90 m 20 m 50 m E

Answer 1

Solution So Etotal at B vez > Ve = 42, 3 m/s (a) Total mechanical energy will remain conserved. Total mechanical energy at& = Toteed mechanical energy at a soox 9,8x90 + x sooxts.o)? 447850 T 447250 (6) sxsoox ve? = 447250 > 1789 42,3 mm)s (C)→ From Conservation of mechanical Evengy Total mechanical energy at D= 447850J 447250 J (d) 1027 N= MXZ-mg = 447250 - Soo X9.8 x to = 20g 250 809 N soox 809 M N+ Mg h 8x500x 2 VD2 Soo X908 15395 N 시 15325

h = 1. (e) Height of Point E from the ground so-40 =lom Kinetic energy of car at E = 447250 500 x9,8X1O 647850-49000 398250 J so k = J 398250 & Kinetic energy of car on the Portion which is Parallel to the ground 398250 Work done by friction -398 ero =k so Wf = J -398250