Final answer:
The force of the floor of the elevator on a 76kg person is 851.2 N. The person feels heavier during the acceleration. It will take approximately 5.35 seconds for the elevator to pass the 5th floor.
Step-by-step explanation:
To determine the force of the floor of the elevator on a 76kg person as it climbs with an acceleration of 1.4 m/s², we use Newton's second law, which states that force equals mass times acceleration (F = m * a). In this case, the person experiences two forces: their weight (force due to gravity) and the force due to the elevator's acceleration.
The weight is calculated by the formula W = m * g, where 'm' is the mass (76kg) and 'g' is the acceleration due to gravity (9.8 m/s²). So, W = 76kg * 9.8 m/s² = 744.8 N (newtons). The additional force due to the elevator's acceleration is F = m * a, with 'a' being the elevator's acceleration (1.4 m/s²). So, F = 76kg * 1.4 m/s² = 106.4 N.
The total force on the person is the sum of their weight and the force due to the elevator's acceleration: 744.8 N + 106.4 N = 851.2 N. Therefore, the force of the floor of the elevator on the person is 851.2 N. During the period of acceleration, the person would feel heavier due to the additional upward force applied by the elevator.
Bonus Question:
To calculate the time it takes for the elevator to reach the 5th floor, assuming it starts from rest, we can use the kinematic equation: d = (1/2)at², where 'd' is the total distance covered, 'a' is the acceleration, and 't' is the time. The 5th floor is at 5 floors * 4m/floor = 20m above the ground. Substituting the given values into the equation and solving for 't', we get:
20m = (1/2) * 1.4 m/s² * t²
20m = 0.7 m/s² * t²
t² = 20m / 0.7 m/s²
t² ≈ 28.57 s²
t ≈ √28.57 s² ≈ 5.35 s
It will take approximately 5.35 seconds for the elevator to pass the 5th floor.