Question

На лёгкой нерастяжимой нити, перекинутой через блок, подвешены два груза одинаковой массы М. Грузы покоятся. На один из грузов устанавливаются перегрузок массой m=300г. Через t=0.8 сек после этого один из грузов оказался выше другого на h=100см. Определите массу М. Массой блока и трением оси можно пренебречь.

Answer 1

устанавливаю перегрузок 
началось движение
Fp=F1-F2=(M+m)g-Mg=mg
Fp=(2M+m)a
приравниваем правые части
mg=(2M+m)a
а=mg/(2M+m) (1)
каждое из тел движется с ускорением а в разные стороны и проходит за 0.8 с
расстояние h/2 запишем уравнение движения h/2=at^2/2 (2)
подставим (1) в (2) и найдем массу М
М=0.81 кг =810 г

Answer 2

Problem Analysis

We have two identical masses, M, suspended from a light, inextensible thread passing over a pulley. One of the masses has an additional load of mass m = 300 g placed on it. After t = 0.8 s, one of the masses is found to be h = 100 cm higher than the other. We need to determine the value of M, given that the mass of the pulley and the friction of the axis can be neglected.

Solution

To solve this problem, we can use the principles of Newton's laws of motion and the concept of equilibrium.

Let's consider the forces acting on the system. The mass M on the left side is being pulled downward by its weight, which can be represented as Mg, where g is the acceleration due to gravity. The mass M on the right side is being pulled upward by the tension in the thread. The additional load of mass m = 300 g on the right side also contributes to the tension in the thread.

At equilibrium, the net force on each mass is zero, and the tension in the thread is constant throughout its length. Therefore, we can write the following equation:

Tension in the thread = Mg + Tension due to additional load

Since the masses are identical, the tension in the thread is the same on both sides. Let's denote the tension in the thread as T.

On the left side: - Weight of mass M: Mg

On the right side: - Weight of mass M: Mg - Weight of additional load m: mg

Using the equation for tension, we can write:

T = Mg + Mg + mg

Simplifying the equation:

T = 2Mg + mg

Now, let's consider the displacement of the masses. The mass M on the left side moves downward by h = 100 cm, while the mass M on the right side moves upward by the same distance. The displacement is related to the time and acceleration by the equation:

h = (1/2)gt^2

Substituting the given values, we have:

0.01 m = (1/2)g(0.8 s)^2

Simplifying the equation:

g = 0.01 m / (0.4 s^2)

Now, we can substitute the value of g into the equation for tension:

T = 2Mg + mg

T = 2M(0.01 m / (0.4 s^2)) + m(0.01 m / (0.4 s^2))

Simplifying the equation:

T = (0.02M + 0.01m) g

Finally, we can substitute the value of tension T into the equation:

T = Mg + Mg + mg

(0.02M + 0.01m) g = Mg + Mg + mg

Simplifying the equation:

0.02Mg + 0.01mg = 2Mg + mg

0.01Mg = 0.01mg

Cancelling out the common factors:

M = m

Therefore, the mass M is equal to the mass of the additional load m, which is 300 g.

Answer

The mass M is equal to the mass of the additional load m, which is 300 g.