A 535 kg roller coaster car began at rest at the top of a 93.0 m hill. Now it is at the top of the first loop-de-loop. This roller coaster s track is nearly frictionless, so resistance can be ignored. [ Using g = 9.8 m/s2, The car has both potential and kinetic energy, and it is moving at 24.6 m/s best describes the roller coaster car when it is at the top of the loop-de-loop. Solution: U=mgh=(535 kg)(9.8 m/s^2)(62.0 m)=325,066 J; K=E-U=487,599 J-325,066 J=162,533 J; K=1/2 mv^2; v=\sqrt 2K/m = sqrt 2. 162,533 J/535 kg =24.6 m/s. ]