Question

A roller-coaster car has a mass of 1200 kg when fully
loaded with passengers. As the car passes over the top of a circular
hill of radius 18 m, its speed is not changing. At the top of the
hill, what are the (a) magnitude FN and (b) direction (up or
down) of the normal force on the car from the track if the car’s
speed is v
11 m/s? What are (c) FN and (d) the direction if v

14 m/s?

Answer 1

At the top of the hill,the car experiences the centripetal force that keeps it in a circular movement and it's weight, so the platform experiences both forces and reflects it as the Normal Force going upwards since Weight goes down and Centripetal force goes to the center of the circle which is the same direction as the weight.

Answer 2

a)
The normal force at the top of the hill is equal to themass*gravity minus the centripetal force Fc
Fn=mg-Fc
Fn=mg-mv2/r
Fn=1200*9.8-1200(11^2)/18
Fn=3693 N
since it is positive it is up.

Answer 3

b) m=1200kg r=18m v=11m/s
F(c)=(mv^2)/r
..=(1200x11^2)/18
..=(1200x121)/18
..=145200/18
..=8066.6666
..=8100N towards the center of the circle, since the force changing the direction of the car's motion is centripetal force. Centripetal force is perpendicular to velocity.

Answer 4

c & d) M=1200kg r=18m v= 14m/s

F(c)=(mv^2)/r
......=(1200x14^2)/18
......=(1200x196)/18
......=(235200)/18
......=13066.666
......=13000N, the direction would be the same as above.

Answer 5

ok @ASAAD123

Answer 6

got it!!

Answer 7

YES