Question

ПОМОГИТЕ ПРОШУ ПОМОГИТЕ ПОЖАЛУЙСТА СРОЧНО Груз подвешен на тросе жёсткостью к = 400 н/м к потолку лифта. Лифт из состояния покоя равноускоренно опускается вниз на расстояние h = 4 м в течение t = 2 с. Удлинение троса при неподвижном относительно лифта грузе равно Дх = 1,5 см? Определите ускорение а, с которым опускается лифт. Ответ выразить в м/с², округлив до целых. Определите силу натяжения троса Т. Ответ выразите в Н, округлив до целых. Определите массу груза м. Ответ выразите в г, округлив до целых.

Answer 1

Ответ:

Ускорение а = 2 м/с².

Сила натяжения троса Т = 4,8 Н.

Масса груза m = 600 г.

Объяснение:

k = 400 H/м

v₀ = 0

h = 4 м

t₁ = 2 c

$d_x = 1.5~cm = 0.015~(m)$

g = 10 м/с²

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a - ? - ускорение

Т - ? - cила натяжения троса

m - масса груза

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Пройденное расстояние

h = v₀t₁ + 0.5at₁²

v₀ = 0

Ускорение

$a = \dfrac{2h}{t_1^2} = \dfrac{2\cdot 4}{2^2} = 2~(m/s^2)$

Статическое удлинение троса

$d_x = \dfrac{mg}{k}$

откуда масса груза

$m = \dfrac{d_x\cdot k}{g} = \dfrac{0.015\cdot 400}{10} = 0.6~(kg) = 600~g$

Cила натяжения троса

T = m · (g - a) = 0.6 · (10 - 2) = 4.8 (H)

Answer 2

Problem Analysis

We are given the following information: - The stiffness of the rope, k, is 400 N/m. - The distance the lift descends, h, is 4 m. - The time taken for the descent, t, is 2 s. - The elongation of the rope when the lift is stationary, Δx, is 1.5 cm.

We need to find: - The acceleration, a, at which the lift descends. - The tension force, T, in the rope. - The mass of the load, m.

Solution

To solve this problem, we can use the concept of Hooke's Law, which states that the force required to stretch or compress a spring is directly proportional to the displacement of the spring from its equilibrium position.

Step 1: Finding the acceleration (a)

We can use the equation of motion for uniformly accelerated linear motion:

h = (1/2) * a * t^2 [[1]]

where h is the distance, a is the acceleration, and t is the time.

Rearranging equation [[1]] to solve for a:

a = (2 * h) / t^2 [[2]]

Substituting the given values:

a = (2 * 4) / (2^2) = 4 / 4 = 1 m/s^2 [[3]]

Therefore, the acceleration of the lift is 1 m/s^2.

Step 2: Finding the tension force (T)

The tension force in the rope can be found using Hooke's Law:

T = k * Δx [[4]]

where T is the tension force, k is the stiffness of the rope, and Δx is the elongation of the rope.

Substituting the given values:

T = 400 N/m * 1.5 cm = 400 N/m * 0.015 m = 6 N [[5]]

Therefore, the tension force in the rope is 6 N.

Step 3: Finding the mass of the load (m)

The tension force in the rope is equal to the weight of the load:

T = m * g [[6]]

where T is the tension force, m is the mass of the load, and g is the acceleration due to gravity.

Rearranging equation [[6]] to solve for m:

m = T / g [[7]]

Substituting the given values:

m = 6 N / 9.8 m/s^2 ≈ 0.61 kg [[8]]

Therefore, the mass of the load is approximately 0.61 kg.

Summary

- The acceleration of the lift is 1 m/s^2. - The tension force in the rope is 6 N. - The mass of the load is approximately 0.61 kg.

Please let me know if I can help you with anything else.