Question

Пассажирский лифт начинает движение из состояния покоя и,двигаясь равноускоренно вниз. за время 6.0с проходит путь S при таком движении лифта вес пассажира массой 60 кг равен 560 Н. ускорение свободного падение равно 10 м/с^2. Определить путь.Напишите решение!

Answer 1

По 2 з. Ньютона mg-Fупр.=ma; Р=Fупр=m(g-a)=560H, 10-a=560/60=9,(3);
а=10-9,3=0,7м/с^2,
S=at^2/2=0,7*36/2=0,7*18=12,6 м.

Answer 2

Problem Analysis

We are given the following information: - The passenger elevator starts moving from rest and accelerates uniformly downwards. - The elevator travels a distance of S in 6.0 seconds. - The weight of the passenger is 560 N, and their mass is 60 kg. - The acceleration due to gravity is 10 m/s^2.

We need to determine the distance traveled by the elevator.

Solution

To solve this problem, we can use the equations of motion for uniformly accelerated linear motion. The relevant equation for this problem is:

S = ut + (1/2)at^2

Where: - S is the distance traveled by the elevator - u is the initial velocity of the elevator (which is 0 since it starts from rest) - a is the acceleration of the elevator - t is the time taken to travel the distance S

We are given that the elevator moves uniformly and accelerates downwards. Therefore, the acceleration of the elevator is equal to the acceleration due to gravity, which is 10 m/s^2.

Substituting the given values into the equation, we have:

S = 0 + (1/2)(10)(6.0)^2

Simplifying the equation, we get:

S = 0 + 300

Therefore, the distance traveled by the elevator is 300 meters.

Answer

The elevator travels a distance of 300 meters.