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 5.70 m/s2. This acceleration only occurs briefly at the beginning of the ride up. Her mass is 88.0 kg. What is the normal force exerted by the floor of the elevator on the student during her brief acceleration

N = 1364 N

Explanation:

given data

accelerate upward = 5.70 m/s²

mass = 88.0 kg

solution

normal force is in upward direction so, weight of the student in downward direction and acceleration is in upward direction so formula is express as

N - mg = ma        1

N = m × (g+a)

put here value

N = 88.0 × (9.8 + 5.70)

N = 1364 N

a displacement distance is size of displ

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