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A 62.0 kg person in a rollercoaster moves 29.2 m/s at the bottom of a curved track of radius 33.7 m. What is the normal force acting on the person? (Unit = N)​

A 62.0 kg person in a rollercoaster moves 29.2 m/s at the bottom of a curved track-example-1
User Happybuddha
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2 Answers

23 votes
23 votes
To find the normal force acting on the person, we can use the equation:

F_normal = m * g * cos(theta) + m * a_t

Where:

F_normal is the normal force acting on the person
m is the mass of the person (62.0 kg)
g is the acceleration due to gravity (9.81 m/s^2)
theta is the angle between the normal force and the vertical direction
a_t is the tangential acceleration of the person

The tangential acceleration of the person can be found using the equation:

a_t = v^2 / r

Where:

a_t is the tangential acceleration of the person
v is the velocity of the person (29.2 m/s)
r is the radius of the curved track (33.7 m)

Substituting these values into the equation, we find that the tangential acceleration of the person is:

a_t = (29.2 m/s)^2 / (33.7 m) = 16.7 m/s^2

Since the person is at the bottom of the curved track, the angle theta is equal to 90 degrees (the normal force is perpendicular to the track). Therefore, the cosine of theta is zero.

Substituting these values into the equation for the normal force, we find that the normal force acting on the person is:

F_normal = (62.0 kg) * (9.81 m/s^2) * (0) + (62.0 kg) * (16.7 m/s^2) = 1030 N

So the normal force acting on the person is approximately 1030 N.
User Phaxian
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3.3k points
14 votes
14 votes

Answer:

2180

Step-by-step explanation:

Worked for Acellus

User Evan Benn
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3.0k points