Question

Шарик, подвешенный на нити, совершает свободные колебания. Полная механическая энергия колеблющегося шарика равна 0,05 Дж. Определите скорость шарика при прохождении им положения равновесия, если его масса равна 0,1 кг. Силой трения в точке подвеса и силой сопротивления воздуха пренебречь.

Answer 1

По формуле кинетической энергии Е = 0,5mv² выразить скорость v = 1 м/с

Answer 2

Calculation of the velocity of the oscillating ball

To determine the velocity of the ball when it passes through the equilibrium position, we can use the principle of conservation of mechanical energy. The total mechanical energy of the oscillating ball is equal to the sum of its kinetic energy and potential energy.

Given: - Total mechanical energy (E) = 0.05 J - Mass of the ball (m) = 0.1 kg

The total mechanical energy (E) can be expressed as the sum of the kinetic energy (K) and potential energy (U):

E = K + U

The kinetic energy of the ball is given by the equation:

K = (1/2)mv^2

where: - m is the mass of the ball - v is the velocity of the ball

The potential energy of the ball is given by the equation:

U = mgh

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 of the ball from the equilibrium position

Since the ball is at its equilibrium position, the height (h) is zero. Therefore, the potential energy (U) is also zero.

Substituting the values into the equation for total mechanical energy:

0.05 J = (1/2)(0.1 kg)v^2 + 0

Simplifying the equation:

0.05 J = (0.05 kg)v^2

Solving for v:

v^2 = (0.05 J) / (0.05 kg)

v^2 = 1 J/kg

Taking the square root of both sides:

v = √(1 J/kg)

Calculating the value:

v ≈ 1 m/s

Therefore, the velocity of the ball when it passes through the equilibrium position is approximately 1 m/s.