Final answer:
The normal force exerted on a coaster at the top of a loop is equal to the sum of the gravitational force and the centripetal force. To find the speed of the roller coaster at the top of the loop, you can use the centripetal force equation. Substituting the given values into the equation will give you the speed of the roller coaster at the top of the loop.
Step-by-step explanation:
The normal force exerted on a coaster at the top of a loop is equal to the sum of the gravitational force and the centripetal force. The normal force is the force exerted by a surface to support the weight of an object resting on it. When a coaster is at the top of a loop, the normal force needs to be greater than the weight of the coaster to provide the necessary centripetal force. In this case, the downward acceleration of the car is given as 1.50 g, where g is the acceleration due to gravity.
To find the speed of the roller coaster at the top of the loop, we can use the centripetal force equation:
F_c = m * a_c
where F_c is the centripetal force, m is the mass of the car, and a_c is the centripetal acceleration. The centripetal force at the top of the loop is provided by the normal force and the gravitational force:
F_c = m * g + m * a
where g is the acceleration due to gravity, and a is the acceleration of the car. In this case, a = 1.50 g.
We can solve for the speed of the roller coaster at the top of the loop by equating the centripetal force to the centripetal acceleration:
m * g + m * a = m * v^2 / r
where v is the speed of the roller coaster at the top of the loop, and r is the radius of curvature. Rearranging the equation gives:
v = sqrt((g + a) * r)
Substituting the given values of g = 9.8 m/s^2, a = 1.50 * 9.8 m/s^2, and r = 15.0 m into the equation gives:
v = sqrt((9.8 + 1.50 * 9.8) * 15.0)
v = sqrt((9.8 * 2.50) * 15.0)
v = sqrt(24.5 * 15.0)
v = sqrt(367.5)
v ≈ 19.17 m/s
Therefore, the speed of the roller coaster at the top of the loop is approximately 19.17 m/s.