Question

Two blocks with different masses are attached to either end of a light rope that passes over a light, frictionless pulley suspended from the ceiling. The masses are released from rest, and the more massive one starts to descend. After this block has descended 1.20 m, its speed is 3.00 m/s. If the total mass of the two blocks is 22.0 kg, what is the mass of each block?

Short answer

The masses are approximately 15.20 kg and 6.80 kg.

Step-by-step solution

Understand the Problem

We have two blocks connected by a rope passing over a pulley. The block with the larger mass descends 1.20 m, achieving a speed of 3.00 m/s. The total mass of the blocks is 22.0 kg. We need to find the mass of each block.

Apply the Law of Conservation of Mechanical Energy

Since the system starts from rest and the pulley is frictionless, we can use conservation of mechanical energy. The potential energy lost by the descending block is converted into kinetic energy. The equation is: \[(m_1 - m_2)gh = \frac{1}{2}(m_1 + m_2)v^2\]where \(g = 9.81\, \text{m/s}^2\), \(h = 1.20\, \text{m}\), \(v = 3.00\, \text{m/s}\).

Express Masses Using Total Mass Constraint

Let \(m_1\) be the mass of the descending block and \(m_2\) the mass of the ascending block. We have \(m_1 + m_2 = 22\, \text{kg}\). We can express \(m_1\) in terms of \(m_2\):\[m_1 = 22 - m_2\]

Substitute Masses into the Energy Equation

Substitute \(m_1 = 22 - m_2\) into the energy equation:\[((22 - m_2) - m_2)gh = \frac{1}{2}(22)v^2\]Simplifying the mass terms gives:\[(22 - 2m_2)gh = 11v^2\]

Solve for the Mass of \(m_2\)

Substitute \(g = 9.81\, \text{m/s}^2\), \(h = 1.20\, \text{m}\), and \(v = 3.00\, \text{m/s}\) into the equation:\[(22 - 2m_2) \cdot 9.81 \cdot 1.20 = 11 \cdot 3.00^2\]Calculate the right side: \(11 \cdot 9 = 99\).Solve for \(m_2\):\[22 - 2m_2 = \frac{99}{9.81 \times 1.20}\]

Calculate Numerical Solution

Calculate \(9.81 \times 1.20 = 11.772\). Now solve:\[22 - 2m_2 = \frac{99}{11.772} \approx 8.405\]\[2m_2 = 22 - 8.405 = 13.595\]\[m_2 \approx \frac{13.595}{2} \approx 6.80\, \text{kg}\]Then, \(m_1 = 22 - m_2 \approx 15.20\, \text{kg}\).

Key concepts

Mechanical Energy Conservation

In a physics problem involving mechanical energy conservation, understanding the transformation of energy within a system is crucial. This principle states that the total mechanical energy (sum of potential and kinetic energy) of an isolated system remains constant, as long as no external forces (like friction) act on it. A practical application of this principle is seen in a block-and-pulley system.
When the blocks are released from rest, the potential energy due to the height of a block is converted into kinetic energy as it descends. This principle is mathematically expressed as:

• Potential energy lost = Kinetic energy gained.

In the provided problem, the gravitational potential energy converted into kinetic energy allows us to solve for unknown variables, such as the masses of the blocks. Understanding how energy is conserved helps simplify complex dynamics into manageable equations.

Frictionless Pulley System

A frictionless pulley system is ideal for understanding basic principles of physics, like mechanical advantage and energy conservation, without the complications introduced by friction. Since friction is absent in our scenario, we can assume that the only forces at play are gravitational and tensions within the system.
This simplification leads to reliable energy transitions between the blocks in the system. The pulley merely changes the direction of the force applied by gravity, causing the heavier block to descend while lifting the lighter one. The lack of friction ensures that no mechanical energy is lost to heat or sound. This setup allows us to focus solely on the relationship between masses, speed, and acceleration, making it a perfect real-world application of theoretical physics principles.

Mass Calculation

When tasked with deriving the individual masses of blocks within a pully system, understanding and utilizing equations effectively is critical. In our exercise, the total mass of both blocks is provided, allowing us to express one mass in terms of the other:

• Let the heavier block be represented as \(m_1\) and the lighter as \(m_2\):
  \(m_1 + m_2 = 22\, \text{kg}\).

Using this initial relationship, we substitute into the equation derived from energy conservation. This equation relates the differences in gravitational potential energy to the resultant kinetic energy. Solving this equation involves algebraic manipulation and substitution to find the individual values:

• A crucial step is calculating the balance between the given total mass and the speed reached after a given descent.

Understanding how to set up and solve these equations gives insight into how mass influences movement in mechanical systems.

Kinematics Equations

Kinematics equations are essential tools in analyzing motion, especially under uniform acceleration like gravity. In this problem, they help confirm the relationship between speed, distance, and time, alongside energy conditions. One essential equation considers the relationship between the speed attained by descending mass:

• Initial speed \(u\): 0 \(\text{m/s}\), as blocks start from rest.
• Final speed \(v\): 3 \(\text{m/s}\), given.
• Distance \(s\): 1.20 \(\text{m}\), over which the mass descends.

Given these parameters, the equations of motion help calculate average acceleration. Acceleration due to gravity (\(g = 9.81 \text{m/s}^2\)) provides the foundation, simplifying problem-solving. These fundamental tools allow the determination of physical quantities crucial in describing rotational and translational movements in funereal and other applications.