A roller coaster, loaded with passenger has a mass of 2000 kg; the radius of curvature of the track at the bottom point of the dip is 24 m. If the vehicle has a speed of 18m/s at this point, what force is exerted on the vehicle by the track? (g = 9.8 m/s2)

To find the force exerted on the roller coaster by the track, we can use the centripetal force formula:

F = (m * v^2) / r

Where:
F is the centripetal force
m is the mass of the roller coaster
v is the velocity of the roller coaster
r is the radius of curvature

Given values:
m = 2000 kg
v = 18 m/s
r = 24 m

To find the force, we'll substitute these values into the formula:

F = (2000 kg * (18 m/s)^2) / 24 m

Now, let's calculate it step by step:

F = (2000 kg * 324 m^2/s^2) / 24 m
F = 648000 kg * m/s^2 / 24 m
F = 27000 kg * m/s^2

Since the unit for force is Newton (N), we can convert kg * m/s^2 to Newton by using the equation: 1 N = 1 kg * m/s^2

F = 27000 N

Therefore, the force exerted on the roller coaster by the track is 27000 Newtons (N).

480000