Asked by A roller-coaster car
A roller-coaster car speeds down a hill past point A where R1 = 8.6 m and then rolls up a hill past point B where R2 = 16.0 m.
(a) The car has a speed of 18.4 m/s at point A. if the track exerts a normal force on the car of 2.21 104 N at this point, what is the mass of the car?
(b) What is the maximum speed the car can have at point B for the gravitational force to hold it on the track?
(a) The car has a speed of 18.4 m/s at point A. if the track exerts a normal force on the car of 2.21 104 N at this point, what is the mass of the car?
(b) What is the maximum speed the car can have at point B for the gravitational force to hold it on the track?
Answers
Answered by
Bot
a) The mass of the car can be calculated using the equation F = ma, where F is the normal force, m is the mass of the car, and a is the acceleration. Rearranging the equation to solve for m, we get m = F/a. Substituting the given values, we get m = 2.21 x 104 N / (18.4 m/s2) = 1.2 x 103 kg.
b) The maximum speed the car can have at point B for the gravitational force to hold it on the track can be calculated using the equation F = mg, where F is the gravitational force, m is the mass of the car, and g is the acceleration due to gravity. Rearranging the equation to solve for v, we get v = √(2Rg), where R is the radius of the hill. Substituting the given values, we get v = √(2 x 16.0 m x 9.8 m/s2) = 20.2 m/s.
b) The maximum speed the car can have at point B for the gravitational force to hold it on the track can be calculated using the equation F = mg, where F is the gravitational force, m is the mass of the car, and g is the acceleration due to gravity. Rearranging the equation to solve for v, we get v = √(2Rg), where R is the radius of the hill. Substituting the given values, we get v = √(2 x 16.0 m x 9.8 m/s2) = 20.2 m/s.