A student takes the elevator up to the fourth floor to see her favorite physics instructor. She stands on the floor of the elevator, which is horizontal. Both the student and the elevator are solid objects, and they both accelerate upward at 5.15 m/s2. This acceleration only occurs briefly at the beginning of the ride up. Her mass is 72.0 kg. What is the normal force exerted by the floor of the elevator on the student during her brief acceleration? (Assume the j direction is upward.)

Question

A student takes the elevator up to the fourth floor to see her favorite physics instructor. She stands on the floor of the elevator, which is horizontal. Both the student and the elevator are solid objects, and they both accelerate upward at 5.15 m/s2. This acceleration only occurs briefly at the beginning of the ride up. Her mass is 72.0 kg. What is the normal force exerted by the floor of the elevator on the student during her brief acceleration? (Assume the j direction is upward.)

Answer 1

force=mg+ma=72(9.8+5.15)

Answer 2

To find the normal force exerted by the floor of the elevator on the student during the acceleration, we need to consider the forces acting on the student and use Newton's second law of motion.

Newton's Second Law: ΣF = ma

In this case, the forces acting on the student are:
1. Gravitational force (weight): mg
2. Normal force: N

Since the elevator and the student are both accelerating upward, the net force in the vertical direction can be defined as follows:

ΣF = N - mg

Using Newton's second law of motion, we can write:

ΣF = ma

Therefore:

N - mg = ma

We can substitute the values given in the question:
m (mass of the student) = 72.0 kg
a (acceleration) = 5.15 m/s^2
g (acceleration due to gravity) = 9.8 m/s^2

N - (72.0 kg) * (9.8 m/s^2) = (72.0 kg) * (5.15 m/s^2)

N - 705.6 N = 370.8 N

To find the normal force, we can add 705.6 N to both sides of the equation:

N = 370.8 N + 705.6 N
N = 1076.4 N

Therefore, the normal force exerted by the floor of the elevator on the student during her brief acceleration is 1076.4 N.

Answer 3

To find the normal force exerted by the floor of the elevator on the student during the brief acceleration, we need to use Newton's second law of motion:

F_net = ma

In this case, the net force acting on the student is the difference between the gravitational force (mg) pulling her downward and the force exerted by the floor (normal force, N) pushing her upward. So, we can rewrite the equation as:

mg - N = ma

Now, let's substitute the given values:

m = 72.0 kg (mass of the student)
g = 9.8 m/s^2 (acceleration due to gravity)
a = 5.15 m/s^2 (acceleration of the elevator)

Plugging in these values:

(72.0 kg)(9.8 m/s^2) - N = (72.0 kg)(5.15 m/s^2)

Simplifying the equation:

705.6 N - N = 369.6 N

Combining the terms with N:

705.6 N - N = 369.6 N
705.6 N - 369.6 N = N
336 N = N

Therefore, the normal force exerted by the floor of the elevator on the student during her brief acceleration is 336 N.