Answer:
The minimum amount that the spring must be compressed if the roller coaster car is to stay on the track is 2.87 meters.
Step-by-step explanation:
The energy required at the top of the roller coaster loop can be determined by considering the energy of the car, which must be supplied at launch by the giant spring.
The car has potential energy due to its height at the top of the track. It has kinetic energy due to its tangential velocity. In order for the car to stay on the track, the tangential velocity must make the required centripetal acceleration equal to the acceleration due to gravity. Where v is the linear velocity and r is the track radius, the relation to gravity is ...
g = v²/r ⇒ v² = rg
Then the kinetic energy at the top of the track is ...
KE(top) = 1/2mv² = 1/2mrg
And the potential energy at the top of the track is ...
PE(top) = mgh = 2mgr . . . . . where h is the height of the car, twice the radius
The total energy of the car at the top of the track is then ...
KE(top) +PE(top) = 1/2mrg +2mrg = 5/2mrg
At the bottom of the track, the energy supplied by the compressed spring will be ...
PE = 1/2kx² . . . . . where k is the spring constant, and x is the amount of compression
For the spring to supply the necessary total energy, we must have ...
1/2kx² = 5/2mrg
x² = (5mrg)/k = 5·(840 kg)(6.2 m)(9.8 m/s²)/(31000 N/m) = 8.232 m²
x = √8.232 m ≈ 2.87 m
The minimum amount that the spring must be compressed if the car is to stay on the track is 2.87 meters.
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Additional comment
The velocity of the car at launch is about 17.43 m/s. Its kinetic energy is about 128 kJ. About 102 kJ of that is converted to potential energy at the top of the track, where the velocity slows to about 7.79 m/s.