Question

Помогите решить задачу пожалуйста.шарик массой 50 г, подвешенный на нерастяжимой нити длиной 40 см отклонили от положения равновесия и отпустили.При прохождении положения равновесия модуль скорости движения шарика 2м/с. Определите модуль силы, с которой шарик действует на нить при прохождении им положения равновесия.

Answer 1

Сначала находим ускорение: а= V-Vо/t. t в данной задаче будет являться периодом колебаний, он вычисляется по формуле T=2п* корень из L/g. Т=1,256с. Подставляем значение t в формулу ускорения(Vо равно нулю, так как сказано, что шарик привели в движение из положения равновесия). Ускорение равно а=1,6 м/с^2. Затем по второму закону Ньютона считаем T-силу натяжения, направленную противоположно силе тяжести. Получаем уравнение: Т-F=ma; Т=ma+F(Сила  тяжести равная mg)=0,58 Н. Ответ: 0,58 Н.

Answer 2

Problem Analysis

We are given a ball of mass 50 g that is suspended from an inextensible string of length 40 cm. The ball is initially displaced from its equilibrium position and then released. When the ball passes through its equilibrium position, its velocity is 2 m/s. We need to determine the magnitude of the force exerted by the ball on the string when it passes through its equilibrium position.

Solution

To solve this problem, we can use the principle of conservation of mechanical energy. The total mechanical energy of the system (ball and string) remains constant throughout the motion.

The total mechanical energy of the system is given by the sum of the potential energy and the kinetic energy:

Total mechanical energy = Potential energy + Kinetic energy

At the highest point of the ball's motion (when it is at its maximum displacement from the equilibrium position), the potential energy is maximum and the kinetic energy is zero. At the equilibrium position, the potential energy is zero and the kinetic energy is maximum.

Since the ball is released from rest at the maximum displacement, the potential energy at the highest point is equal to the potential energy at the equilibrium position. Therefore, we can equate the potential energy at the highest point to the kinetic energy at the equilibrium position:

Potential energy at highest point = Kinetic energy at equilibrium position

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

Potential energy = mgh

where m is the mass of the ball, g is the acceleration due to gravity, and h is the height of the ball from the equilibrium position.

The kinetic energy of the ball at the equilibrium position is given by the formula:

Kinetic energy = (1/2)mv^2

where m is the mass of the ball and v is the velocity of the ball at the equilibrium position.

By equating the potential energy at the highest point to the kinetic energy at the equilibrium position, we can solve for the height h:

mgh = (1/2)mv^2

Simplifying the equation, we get:

gh = (1/2)v^2

Now, we can substitute the given values into the equation and solve for h:

g = 9.8 m/s^2 (acceleration due to gravity) v = 2 m/s (velocity at equilibrium position)

h = (1/2)(v^2)/g

Substituting the values, we get:

h = (1/2)(2^2)/9.8 = 0.204 m

Now that we have the height h, we can calculate the force exerted by the ball on the string at the equilibrium position. The force is given by the formula:

Force = mass x acceleration

The acceleration of the ball at the equilibrium position is given by the centripetal acceleration formula:

Acceleration = v^2 / r

where v is the velocity of the ball at the equilibrium position and r is the radius of the circular path.

In this case, the radius of the circular path is equal to the length of the string, which is 40 cm or 0.4 m.

Substituting the values, we get:

Acceleration = (2^2) / 0.4 = 10 m/s^2

Now, we can calculate the force:

Force = mass x acceleration = 0.05 kg x 10 m/s^2 = 0.5 N

Therefore, the magnitude of the force with which the ball acts on the string at the equilibrium position is 0.5 N.

Answer

The magnitude of the force with which the ball acts on the string at the equilibrium position is 0.5 N.