Question

Небольшой кубик массой m= 1 кг начинает соскальзывать с высоты Н = 3 м по гладкой горке, переходящей в мёртвую петлю радиусом R=1,5 м. С какой силой брусок давит на стенку петли на высоте h=2 м от нижней точки петли? Сделайте рисунок с указанием сил, поясняющий решение. помогите пожалуйста!!!! 50 баллов

Answer 1

Задание 28. Небольшой брусок массой m = 1 кг начинает соскальзывать с высоты Н = 3 м по гладкой горке, переходящей в мёртвую петлю (см. рисунок). Определите радиус петли R, если на высоте h = 2,5 м от нижней точки петли брусок давит на её стенку с силой F = 4 Н. Сделайте рисунок с указанием сил, поясняющий решение.



Решение.

Пусть скорость бруска на высоте h равна v, а в нижней точке петли потенциальная энергия равна нулю. Тогда по закону сохранения механической энергии, получаем:

,

откуда

. (1)

Answer 2

Problem Statement

A small cube with a mass of m = 1 kg starts sliding down a smooth slope from a height of H = 3 m. The slope then transitions into a vertical loop with a radius of R = 1.5 m. What is the force exerted by the cube on the wall of the loop at a height of h = 2 m from the bottom of the loop?

Solution

To solve this problem, we need to consider the forces acting on the cube at the given height h.

1. At the top of the slope (height H = 3 m), the only force acting on the cube is its weight, which can be calculated using the formula:

Weight = mass x gravitational acceleration

Given that the mass of the cube is m = 1 kg and the gravitational acceleration is approximately 9.8 m/s^2, we can calculate the weight:

Weight = 1 kg x 9.8 m/s^2 = 9.8 N 2. As the cube slides down the slope, it gains kinetic energy. At the bottom of the slope, all of the potential energy is converted into kinetic energy. The kinetic energy can be calculated using the formula:

Kinetic Energy = (1/2) x mass x velocity^2

Since the cube starts from rest at the top of the slope, its initial velocity is 0. At the bottom of the slope, the potential energy is completely converted into kinetic energy:

Potential Energy at H = Kinetic Energy at the bottom

m x g x H = (1/2) x m x v^2

Solving for v, we find:

v = sqrt(2 x g x H)

Substituting the values, we get:

v = sqrt(2 x 9.8 m/s^2 x 3 m) = 7.67 m/s 3. As the cube enters the loop, it experiences a centripetal force that keeps it moving in a circular path. At any point in the loop, the net force acting on the cube is the difference between the normal force and the gravitational force.

At the bottom of the loop, the normal force is equal to the sum of the weight and the centripetal force:

Normal Force at the bottom = Weight + Centripetal Force

The centripetal force can be calculated using the formula:

Centripetal Force = (mass x velocity^2) / radius

Substituting the values, we get:

Centripetal Force = (1 kg x (7.67 m/s)^2) / 1.5 m = 39.8 N 4. At a height h = 2 m from the bottom of the loop, the normal force will be different. To find the normal force at this height, we need to consider the forces acting on the cube.

- The weight of the cube is acting vertically downward with a magnitude of 9.8 N. - The centripetal force is acting horizontally inward with a magnitude of 39.8 N.

At the given height h, the normal force can be calculated as the vector sum of the weight and the centripetal force:

Normal Force at height h = sqrt(Weight^2 + Centripetal Force^2)

Substituting the values, we get:

Normal Force at height h = sqrt((9.8 N)^2 + (39.8 N)^2) = 41.1 N Therefore, the force exerted by the cube on the wall of the loop at a height of h = 2 m from the bottom of the loop is approximately 41.1 N.

Diagram

Here is a diagram illustrating the forces acting on the cube at the given height h:

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