A roller-coaster car has a mass of 1240 kg when fully loaded with passengers. As the car passes over the top of a circular hill of radius 22 m, its speed is not changing. (a) At the top of the hill, what is the normal force (using the negative sign for the downward direction) FN on the car from the track if the car's speed is v = 8.7 m/s? (b) What is FN if v = 20 m/s? Use g=9.81 m/s2.

a)      N = 7.90 10³ N,  b) N = -1.04 10⁴ N

Explanation:

a) For this exercise we can use Newton's second law

             N -W = m (-a)

The relationship is centripetal, the negative sign of the acceleration is because it points towards the center of the circle

              a = v² / r

we substitute

             N = mg -m v² /r  

             N = m (g - v² /r)

let's calculate

v = 8.7 m / s

             N = 1240 (9.81 - 8.7²/22)

             N = 7.90 10³ N

b) v = 20 m / s

             N = 1240 (9.81 - 20²/22)

             N = -1.04 10⁴ N

graphing motion: mastery test

209be

what is the instantaneous acceleration of

the particle at point b?

0 meters/second?

10 meters/second?

12 meters/second?

velocity imetersisecond

15 meters/second?

20

time (seconds)

answer: 0 meters/second^2

neither accurate nor precise.