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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

Expert verified
The masses are approximately 15.20 kg and 6.80 kg.

Step by step solution

01

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.
02

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 \(g = 9.81\, \text{m/s}^2\), \(h = 1.20\, \text{m}\), \(v = 3.00\, \text{m/s}\).
03

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 \(m_2\):\[m_1 = 22 - m_2\]
04

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:\[(22 - 2m_2)gh = 11v^2\]
05

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 \(m_2\):\[22 - 2m_2 = \frac{99}{9.81 \times 1.20}\]
06

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, \(m_1 = 22 - m_2 \approx 15.20\, \text{kg}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A cutting tool under microprocessor control has several forces acting on it. One force is \(\overrightarrow{F}\) \(= - \alpha xy^2 \hat\jmath\), a force in the negative \(y\)-direction whose magnitude depends on the position of the tool. For \(a =\) 2.50 N/m\(^3\), consider the displacement of the tool from the origin to the point (\(x =\) 3.00 m, \(y =\) 3.00 m). (a) Calculate the work done on the tool by \(\overrightarrow{F}\) if this displacement is along the straight line \(y = x\) that connects these two points. (b) Calculate the work done on the tool by \(\overrightarrow{F}\) if the tool is first moved out along the \(x\)-axis to the point (\(x =\) 3.00 m, \(y =\) 0) and then moved parallel to the y-axis to the point (\(x =\) 3.00 m, \(y =\) 3.00 m). (c) Compare the work done by \(\overrightarrow{F}\) along these two paths. Is \(\overrightarrow{F}\) conservative or nonconservative? Explain.

A crate of mass \(M\) starts from rest at the top of a frictionless ramp inclined at an angle \(\alpha\) above the horizontal. Find its speed at the bottom of the ramp, a distance \(d\) from where it started. Do this in two ways: Take the level at which the potential energy is zero to be (a) at the bottom of the ramp with \(y\) positive upward, and (b) at the top of the ramp with y positive upward. (c) Why didn't the normal force enter into your solution?

A small block with mass 0.0500 kg slides in a vertical circle of radius \(R =\) 0.800 m on the inside of a circular track. There is no friction between the track and the block. At the bottom of the block's path, the normal force the track exerts on the block has magnitude 3.40 N. What is the magnitude of the normal force that the track exerts on the block when it is at the top of its path?

A small block with mass 0.0400 kg slides in a vertical circle of radius \(R =\) 0.500 m on the inside of a circular track. During one of the revolutions of the block, when the block is at the bottom of its path, point \(A\), the normal force exerted on the block by the track has magnitude 3.95 N. In this same revolution, when the block reaches the top of its path, point \(B\), the normal force exerted on the block has magnitude 0.680 N. How much work is done on the block by friction during the motion of the block from point \(A\) to point \(B\)?

A slingshot will shoot a 10-g pebble 22.0 m straight up. (a) How much potential energy is stored in the slingshot's rubber band? (b) With the same potential energy stored in the rubber band, how high can the slingshot shoot a 25-g pebble? (c) What physical effects did you ignore in solving this problem?

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