This question is incomplete, the complete question is;
The loaded car of a roller coaster has mass M = 320 kg. It goes over the highest hill with a speed v of 21.4 m/s. The radius of curvature R of the hill is 15.8 m.
(a) What is the force (N) that the track must exert on the car? (positive is up)
(b) What must be the force (N) that the car exerts on a 61 kg passenger?
Answer:
a) the force (N) that the track must exert on the car is -6139.14 N
b) the force (N) that the car exerts on a 61 kg passenger is -1170.27 N
Explanation:
Given the data in the question;
Let N represent the force that the track must exerted on the car
Net force on the car Fnet = Mg + N
so
M × a = Mg + N
N = Ma - Mg
N = Ma - M(v²/R)
we substitute
N = (320kg × 9.8m/s²) - ( 320 × ((21.4m/s)² / 15.8 m) )
N = 3136 - ( 320 × 28.9848 )
N = 3136 - 9275.136
N = -6139.14 N
Therefore, the force (N) that the track must exert on the car is -6139.14 N
b) What must be the force (N) that the car exerts on a 61 kg passenger?
Let N represent the force that the car exerts on 61kg passengers
so
Net force of passengers Fnet = mg + N
Ma = Mg + N
N = Ma - Mg
N = Ma - M(v²/R)
N = (61kg × 9.8m/s²) - ( 61 × ((21.4m/s)² / 15.8 m) )
N = 597.8 - ( 61 × 28.9848)
N = 597.8 - 1768.0728
N = -1170.27 N
Therefore, the force (N) that the car exerts on a 61 kg passenger is -1170.27 N
The centripetal force of the track on the car moving in the circular path is [tex]1.465 \times 10^6 \ N[/tex].
The force (N) that the car exerts on a 61 kg passenger is 597.8 N.
The centripetal force of the track on the car moving in the circular path is calculated as follows;
[tex]F_c = \frac{mv^2}{r}\\\\ F_c = \frac{320 \times 21.4^2}{0.1} \\\\F_c = 1.465 \times 10^6 \ N[/tex]
The force (N) that the car exerts on a 61 kg passenger is equal to the force the passenger exerts on the car based on Newton's third law of motion.
F = mg
F = 61 x 9.8
F = 597.8 N
Learn more about centripetal force here: https://brainly.com/question/20905151
