There is a 65 kg student at point A on an amusement park ride. The ride spins the student in a horizontal circle of radius 2.5 meters, at a constant speed of 8.6 meters per second. The floor is lowered and the student remains against the wall without falling to the floor. Approximately what is the magnitude of the centripetal force acting on the student at point A
The magnitude of the centripetal force acting on the student at point A can be calculated using the formula:
F = m * v^2 / r
where F is the centripetal force, m is the mass of the student (65 kg), v is the velocity of the student (8.6 m/s), and r is the radius of the circular path (2.5 m).
Plugging in the values into the formula:
F = 65 kg * (8.6 m/s)^2 / 2.5 m
Calculating this expression gives:
F = 65 kg * 74.76 m^2/s^2 / 2.5 m
= 65 kg * 29.904 m^2/s^2
So, the magnitude of the centripetal force acting on the student at point A is approximately 1943.26 Newtons (N).