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A roller coaster, loaded with passengers, has a mass of 2 000 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 18 m/s at this point, what force is exerted on the vehicle by the track? (g = 9.8 m/s2)

Respuesta :

MathPhys

Answer:

46600 N

Explanation:

The vehicle has two forces acting on it: weight pulling down (away from the center of the track) and normal force of the track pushing up (towards the center of the track).

Sum of the forces towards the center of the track:

∑F = ma

N − mg = m v² / r

N = mg + m v² / r

N = m (g + v² / r)

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

N = 2000 (9.8 + 18² / 24)

N = 46600 N

The track pushes on the vehicle with a force of 46600 Newtons.