A roller-coaster car at a Gold Coast theme park has a mass of 500 kg when fully loaded with passengers. At the bottom of a circular dip of radius 40 m (as shown in the figure) the car has a speed of 16 m/s. What is the magnitude of the force of the track on the car at the bottom of the dip?

Question

Quizplus · Accepted Answer

We can use the centripetal force formula to solve this problem, which is $F_c = \frac{mv^2}{r}$, where $m$ is the mass of the roller-coaster car, $v$ is its velocity, and $r$ is the radius of the circular dip. Plugging in the given values, we get:

$F_c = \frac{(500	ext{ kg})(16	ext{ m/s})^2}{40	ext{ m}} = 1.28	imes 10^4	ext{ N}$

However, this is not the force of the track on the car, because there are other forces acting on the car. The weight of the car and its passengers, which is given by $mg$, acts downward, and the normal force of the track, which is perpendicular to the track, acts upward. Since the car is in equilibrium at the bottom of the dip, the normal force must equal the sum of the weight and the centripetal force:

$F_{	ext{net}} = F_n - mg = F_c$

$F_n = F_c + mg = (1.28	imes 10^4	ext{ N}) + (500	ext{ kg})(9.81	ext{ m/s}^2) = 1.80	imes 10^4	ext{ N}$

Therefore, the magnitude of the force of the track on the car at the bottom of the dip is $1.80	imes 10^4	ext{ N}$, which is equivalent to 8.1 kN (kilonewtons). The answer is B.

A Roller-Coaster Car at a Gold Coast Theme Park Has