Question

шарик массой 100г совершает колебание на нити.найдите скорость и кинетическую энергию шарика в точке равновесия,если центр тяжести шарика при его максимальном отклонении находится выше точки равновесия на 2,5 см.​

Answer 1

Ответ:

см фото.

Объяснение:

Answer 2

Calculation of Speed and Kinetic Energy of the Ball

To find the speed and kinetic energy of the ball at the point of equilibrium, we can use the principle of conservation of mechanical energy. The mechanical energy of the ball is conserved throughout its motion.

The mechanical energy of the ball consists of two components: potential energy and kinetic energy. At the point of maximum displacement, the ball has maximum potential energy and zero kinetic energy. At the point of equilibrium, the ball has maximum kinetic energy and zero potential energy.

Let's denote the mass of the ball as m (100g = 0.1kg) and the maximum displacement as l (2.5cm = 0.025m).

The potential energy of the ball at the point of maximum displacement is given by the formula:

Potential Energy = m * g * h

Where: - m is the mass of the ball - g is the acceleration due to gravity (approximately 9.8 m/s^2) - h is the height difference between the point of maximum displacement and the point of equilibrium (2.5cm = 0.025m)

Substituting the values, we get:

Potential Energy = 0.1kg * 9.8 m/s^2 * 0.025m

To find the kinetic energy at the point of equilibrium, we can use the conservation of mechanical energy:

Potential Energy = Kinetic Energy

Therefore, the kinetic energy of the ball at the point of equilibrium is:

Kinetic Energy = 0.1kg * 9.8 m/s^2 * 0.025m

Now, let's calculate the speed of the ball at the point of equilibrium. The kinetic energy can also be expressed as:

Kinetic Energy = (1/2) * m * v^2

Where: - m is the mass of the ball - v is the speed of the ball

Equating the two expressions for kinetic energy, we have:

(1/2) * m * v^2 = 0.1kg * 9.8 m/s^2 * 0.025m

Simplifying the equation, we find:

v^2 = 2 * 9.8 m/s^2 * 0.025m

Taking the square root of both sides, we get:

v = sqrt(2 * 9.8 m/s^2 * 0.025m)

Now, let's calculate the values.

Calculation:

Using the given values: - Mass of the ball (m) = 0.1 kg - Maximum displacement (l) = 0.025 m - Acceleration due to gravity (g) = 9.8 m/s^2

We can calculate the potential energy and kinetic energy of the ball at the point of equilibrium:

Potential Energy = 0.1 kg * 9.8 m/s^2 * 0.025 m

Kinetic Energy = 0.1 kg * 9.8 m/s^2 * 0.025 m

To find the speed of the ball at the point of equilibrium, we can use the equation:

v = sqrt(2 * 9.8 m/s^2 * 0.025 m)

Now, let's calculate the values:

Potential Energy = 0.1 kg * 9.8 m/s^2 * 0.025 m = 0.0245 J

Kinetic Energy = 0.1 kg * 9.8 m/s^2 * 0.025 m = 0.0245 J

v = sqrt(2 * 9.8 m/s^2 * 0.025 m) ≈ 0.7 m/s

Therefore, the speed of the ball at the point of equilibrium is approximately 0.7 m/s and the kinetic energy of the ball at the point of equilibrium is approximately 0.0245 J.

Please note that these calculations assume ideal conditions and neglect any frictional forces or air resistance that may affect the actual values.