A student with mass (72.0) kg rides on a roller coaster with a vertical loop. At the top of the loop the track has a radius of (23.7) m and the roller coaster moves at speed of (25.9) m/s at the top. Determine the normal force from the seat on the student at the top of the loop. Give your answer in newtons (N) and with 3 significant figures.
force at top=mg-mv^2/r If it is negative, she better be holding on.
To solve this problem, we need to consider the forces acting on the student at the top of the loop. The two main forces are the gravitational force (weight) and the normal force from the seat.
1. First, let's calculate the gravitational force acting on the student. The weight, W, is given by the formula:
W = m * g
where m is the mass of the student and g is the acceleration due to gravity (approximately 9.8 m/s²).
Plugging in the values, we have:
W = (72.0 kg) * (9.8 m/s²)
W ≈ 705.6 N (to 3 significant figures)
2. Next, let's analyze the forces at the top of the loop. At the top, the normal force and the weight add up to provide the centripetal force required for circular motion.
The net force must be directed towards the center of the loop. So, we have:
N + W = m * v² / R
where N is the normal force, m is the mass, v is the velocity, and R is the radius of the loop.
Plugging in the given values and solving for N, we get:
N = (m * v² / R) - W
N = (72.0 kg * (25.9 m/s)² / 23.7 m) - 705.6 N
Calculating this further, we have:
N ≈ 309.7 N (to 3 significant figures)
Therefore, the normal force from the seat on the student at the top of the loop is approximately 309.7 N.