A500 kg rollercoaster car starts from rest at the top of a 10.0 m tall hill. it then travels down the track and up a loop. the top of the loop is 7.0 m from the ground. what is the speed of the rollercoaster car at the top of the loop assuming negligible friction losses?

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