Question

брусок масою 2 кг тягнуть рівномірно за допомогою пружини жорсткістю 200 ньютонів на метр на скільки довшою стане пружина якщо коефіцієнт тертя між бруском і столом 0,3 Поможіть будь ласка СРОЧНО​

Answer 1

Ответ:

За законом Гука, сила, яку розвиває пружина, пропорційна подовженню пружини, тобто

F = kx,

де F - сила, k - жорсткість пружини і x - подовження пружини.

Сила тертя між бруском і столом дорівнює

Fт = μmg,

де μ - коефіцієнт тертя, m - маса бруска і g - прискорення вільного падіння.

Оскільки брусок тягнуть рівномірно, сила тертя дорівнює силі пружини, тобто:

Fт = F

μmg = kxx = μmg/k

Підставляючи дані, отримуємо:

x = (0.3)(2 кг)(9.8 м/с^2)/(200 Н/м) = 0.0294 метрів

Отже, пружина подовжиться на 0,0294 метра (або 2,94 сантиметра).

Объяснение:

Answer 2

Problem Analysis

We are given a wooden block with a mass of 2 kg that is being pulled evenly by a spring with a stiffness of 200 N/m. We need to determine how much longer the spring will become if the coefficient of friction between the block and the table is 0.3.

Solution

To solve this problem, we can use Hooke's law and consider the forces acting on the block.

1. First, let's calculate the force exerted by the spring on the block using Hooke's law. Hooke's law states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. The formula for Hooke's law is:

F = k * x

Where: - F is the force exerted by the spring (in newtons) - k is the stiffness of the spring (in newtons per meter) - x is the displacement of the spring from its equilibrium position (in meters)

In this case, the stiffness of the spring is 200 N/m, and we need to find the displacement of the spring. Since the block is being pulled evenly, the displacement of the spring will be equal to the elongation of the spring.

2. Next, let's calculate the force of friction acting on the block. The force of friction can be calculated using the formula:

F_friction = μ * F_normal

Where: - F_friction is the force of friction (in newtons) - μ is the coefficient of friction - F_normal is the normal force (in newtons)

In this case, the normal force is equal to the weight of the block, which can be calculated using the formula:

F_normal = m * g

Where: - m is the mass of the block (in kilograms) - g is the acceleration due to gravity (approximately 9.8 m/s^2)

3. Now, let's calculate the elongation of the spring. Since the block is being pulled evenly, the force exerted by the spring is equal to the force of friction. Therefore, we can equate the two forces:

k * x = μ * F_normal

Substituting the values we have:

200 * x = 0.3 * (2 * 9.8)

Solving for x:

x = (0.3 * 2 * 9.8) / 200

4. Finally, let's calculate how much longer the spring will become. The elongation of the spring is equal to the displacement of the spring, which we calculated in the previous step. Therefore, the spring will become 0.0294 meters (or 29.4 millimeters) longer.

Answer

The spring will become approximately 0.0294 meters (or 29.4 millimeters) longer when pulling a 2 kg wooden block evenly with a spring stiffness of 200 N/m and a coefficient of friction of 0.3 between the block and the table.