A 2000 kg roller coaster is at the top of a loop with a radius of 24 m. If its speed is 18 m/s at this point, what force does it exert on the track

Question

asked Apr 9, 2022 90.5k views

1 Answer

Moledet · Answer 1 · 2022-04-15T19:01:24+0000

Answer:

$46620\ \text{N}$

Step-by-step explanation:

m = Mass of roller coaster = 2000 kg

r = Radius of loop = 24 m

v = Velocity of roller coaster = 18 m/s

g = Acceleration due to gravity =
$9.81\ \text{m/s}^2$

Normal force at the point will be

$N-mg=(mv^2)/(r)\\\Rightarrow N=(mv^2)/(r)+mg\\\Rightarrow N=(2000* 18^2)/(24)+2000* 9.81\\\Rightarrow N=46620\ \text{N}$

The force exerted on the track is
$46620\ \text{N}$ .