A 414 kg rollercoaster car is traveling around a loop that has a radius of 38.5 meters. The track can provide a normal force up to 18500 N. (a) What is the maximum speed the car can be traveling at the bottom of the loop

answer:

Speed of the roller coaster at the top of the loop= 7.67 m/s

Explanation:

using the law of conservation of energy

KEi + PEi= KEf + PEf

KEi= kinetic energy at the top of the hill=0 because the car is at rest there.

PEi= potential energy at the top of the hill

PEf= potential energy at the top of the loop

KEf= kinetic energy at the top of the loop

Also kinetic energy= 1/ 2m v² and potential energy= mgh

m= mass

h= height

v= velocity

so 0+ mghi = 1/2mv² + mg h

500 (9.8)(10)+ 1/2 (500) v²= 500 ( 9.8) (7)

49000+250 v²= 34300

250v²= 14700

v²=58.8

v=7.67 m/s

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