Question

Через блок перекинута невесомая нерастяжимая нить, к концам которой подвешены два груза массами 0,8 кг и 0,2 кг. В начальный момент времени блок зафиксирован и система покоится: оба груза находятся на высоте 1 м над уровнем пола. Затем блок освобождают, и он начинает вращаться без трения. Какую скорость будет иметь первый груз в момент времени, когда он достигнет пола? Сопротивлением пренебречь. Ответ записать в метрах в секунду с точностью до десятых.

Answer 1

неподвижныйм блок не дает выигрыша в силе , поэтому   mg

Answer 2

Problem Analysis

We have a system consisting of two masses, 0.8 kg and 0.2 kg, connected by an inextensible and weightless string. The system is initially at rest, with both masses suspended at a height of 1 m above the floor. The block is then released, and it starts rotating without friction. We need to determine the velocity of the first mass when it reaches the floor.

Solution

To solve this problem, we can apply the principle of conservation of mechanical energy. At the initial moment, the system has potential energy due to the height of the masses. When the first mass reaches the floor, all of its potential energy will be converted into kinetic energy.

The potential energy of an object at a height h is given by the formula: PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.

The kinetic energy of an object with mass m and velocity v is given by the formula: KE = (1/2)mv^2.

Since the system is initially at rest, the total mechanical energy of the system is equal to the potential energy of the masses. When the first mass reaches the floor, all of this potential energy is converted into kinetic energy.

Let's calculate the potential energy of the system at the initial moment: - The potential energy of the first mass is PE1 = m1 * g * h, where m1 = 0.8 kg and h = 1 m. - The potential energy of the second mass is PE2 = m2 * g * h, where m2 = 0.2 kg and h = 1 m.

The total potential energy of the system is the sum of the potential energies of the two masses: PE_total = PE1 + PE2.

Now, let's calculate the velocity of the first mass when it reaches the floor by equating the potential energy to the kinetic energy: PE_total = KE1 PE1 + PE2 = (1/2) * m1 * v^2

We can solve this equation for v to find the velocity of the first mass when it reaches the floor.

Calculation

Let's substitute the given values into the equation and solve for v:

PE1 + PE2 = (1/2) * m1 * v^2

PE1 = m1 * g * h = 0.8 kg * 9.8 m/s^2 * 1 m = 7.84 J

PE2 = m2 * g * h = 0.2 kg * 9.8 m/s^2 * 1 m = 1.96 J

PE_total = PE1 + PE2 = 7.84 J + 1.96 J = 9.8 J

9.8 J = (1/2) * m1 * v^2

Solving for v:

v^2 = (2 * PE_total) / m1

v^2 = (2 * 9.8 J) / 0.8 kg

v^2 = 24.5 m^2/s^2

Taking the square root of both sides:

v = sqrt(24.5) m/s

v ≈ 4.95 m/s

Therefore, the velocity of the first mass when it reaches the floor is approximately 4.95 m/s.

Answer

The velocity of the first mass when it reaches the floor is approximately 4.95 m/s.