Question

Школьник массой 50 кг едет в лифте наверх. скорость лифта равномерно изменятеся от 5 м/с до 1 м/с за 5 секунд. с какой силой школьник давит при этом на пол.лифта?

Answer 1

m=50    v1=5   v2=1    t=5    F=?

a=(v2-v1)/t=(1-5)/5=-0.8

F=m*(g-a)=50*(9.8-0.8)=450 H

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Answer 2

Calculation of the Force Exerted by the Student on the Elevator Floor

To calculate the force exerted by the student on the elevator floor, we need to consider the change in velocity and the mass of the student.

Given: - Mass of the student (m) = 50 kg - Initial velocity of the elevator (v1) = 5 m/s - Final velocity of the elevator (v2) = 1 m/s - Time taken for the change in velocity (t) = 5 seconds

To calculate the force (F), we can use Newton's second law of motion, which states that force is equal to mass multiplied by acceleration:

F = m * a

Acceleration (a) can be calculated using the formula:

a = (v2 - v1) / t

Let's calculate the force exerted by the student on the elevator floor using the given values:

Step 1: Calculate the acceleration (a): a = (1 m/s - 5 m/s) / 5 s = -4 m/s²

Step 2: Calculate the force (F): F = m * a = 50 kg * (-4 m/s²) = -200 N

The negative sign indicates that the force is directed downwards, opposing the motion of the elevator.

Therefore, the student exerts a force of 200 N on the elevator floor in the downward direction.