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Chapter 7: Problem 40

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 \(1.20 \mathrm{m},\) its speed is 3.00 \(\mathrm{m} / \mathrm{s} .\) If the total mass of the two blocks is 15.0 \(\mathrm{kg},\) what is the mass of each block?

Short Answer

Expert verified
Masses are approximately 5.74 kg and 9.26 kg.

Step by step solution

01

Understand the problem

Two blocks with a total mass of 15.0 kg are connected over a frictionless pulley. The goal is to determine the individual masses of the blocks given that the more massive block descends 1.20 m attaining a speed of 3.00 m/s.
02

Set up the equations

Use the principle of energy conservation. The potential energy lost by the heavier block is converted into kinetic energy shared by both blocks:\[ m_1gh = \frac{1}{2}m_1v^2 + \frac{1}{2}m_2v^2 \]where \( m_1 \) is the heavier block, \( m_2 \) is the lighter block, \( g = 9.81 \, \text{m/s}^2 \), \( h = 1.20 \, \text{m} \), and \( v = 3.00 \, \text{m/s} \).
03

Substitute known values

Substitute the known values into the equation:\[ m_1 \times 9.81 \times 1.20 = \frac{1}{2} (m_1 + m_2) \times (3.00)^2 \]Simplified:\[ 11.772\,m_1 = 4.5 (m_1 + m_2) \]
04

Use total mass condition

We know \( m_1 + m_2 = 15.0 \; \text{kg} \). Use this to express \( m_2 \) in terms of \( m_1 \):\[ m_2 = 15.0 - m_1 \]
05

Solve the system of equations

Substitute \( m_2 = 15.0 - m_1 \) into the equation from Step 3:\[ 11.772\,m_1 = 4.5 \times (m_1 + 15.0 - m_1) \]\[ 11.772\,m_1 = 67.5 \]Solve for \( m_1 \):\[ m_1 = \frac{67.5}{11.772} \approx 5.74 \; \text{kg} \]
06

Determine the lighter mass

Use \( m_2 = 15.0 - m_1 \) to find \( m_2 \):\[ m_2 = 15.0 - 5.74 \approx 9.26 \; \text{kg} \]
07

Verify the solution

Check that the energy conversion is consistent with both kinetic and potential energies:\[ m_1gh = \frac{1}{2}(m_1+m_2)v^2 \]Calculate:\[ 5.74 \times 9.81 \times 1.20 \approx \frac{1}{2} \times 15.0 \times (3.00)^2 \]Both sides should roughly balance, confirming the physics is consistent.

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Key Concepts

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

Newton's Laws and Their Role
Newton's laws of motion are fundamental principles that describe the relationships between a body and the forces acting upon it. In this exercise, we are mainly concerned with Newton's Second Law, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Using the equation, \( F = ma \), we can determine the acceleration of the system.

For the two blocks in the pulley system, the force causing motion is the gravitational force acting on the more massive block. According to Newton's Third Law, the tension in the rope is equal and opposite on both blocks, but since the pulley is frictionless, it doesn't affect their motion. Understanding these principles helps determine how forces interact in the system, thus allowing us to find the acceleration and speeds of the blocks.
  • Newton's First Law: Inertia - An object remains at rest or in uniform motion unless acted upon by an external force.
  • Newton's Second Law: Acceleration - \( F = ma \).
  • Newton's Third Law: Action and Reaction - Forces occur in equal and opposite pairs.
Understanding Kinematics in Pulley Systems
Kinematics refers to the study of motion without considering the forces that cause it. In this problem, we explore how the kinematic equations help us calculate the motion of the blocks over time.

The exercise involves two blocks moving in opposite directions as they are connected over a frictionless pulley. After the heavier block descends by 1.20 m, it reaches a speed of 3.00 m/s. This displacement, time, and speed allow us to use kinematic equations to further analyze motion. For example, the equation \( v^2 = u^2 + 2as \) (where \( v \) is final velocity, \( u \) is initial velocity, \( a \) is acceleration, and \( s \) is displacement) can help deduce the acceleration in this arrangement.
  • Initial velocity \( (u) = 0 \), since blocks start from rest.
  • Displacement \( (s) = 1.20 \, m \).
  • Final velocity \( (v) = 3.00 \, m/s \).
Exploring Frictionless Pulley Systems
Frictionless pulley systems are theoretical models used to simplify the analysis of motion in systems that involve ropes and pulleys. In these systems, the pulley is considered to have negligible mass, and there is no friction between the rope and pulley. This assumption allows us to focus on the motion without complicating factors.

In the given exercise, the frictionless nature of the pulley ensures that the tension is uniform across the rope. This means that while the mass of the blocks will affect the acceleration, it won't be reduced by frictional forces. Because the pulley is also light, its mass doesn't influence the calculations of the system's dynamics.
  • Frictionless pulley: Enables uniform tension.
  • Light rope and pulley: Neglects additional mass influence.
  • The focus remains on gravitational forces and masses' inertia.

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Most popular questions from this chapter

CALE A certain spring is found not to obey Hooke's law; it exerts a restoring force \(F_{x}(x)=-\alpha x-\beta x^{2}\) if it is stretched or compressed, where \(\alpha=60.0 \mathrm{N} / \mathrm{m}\) and \(\beta=18.0 \mathrm{N} / \mathrm{m}^{2} .\) The mass of the spring is negligible, (a) Calculate the potential-energy function \(U(x)\) for this spring. Let \(U=0\) when \(x=0 .\) (b) An object with mass 0.900 \(\mathrm{kg}\) on a frictionless, horizontal surface is attached to this spring, pulled a distance 1.00 \(\mathrm{m}\) to the right (the \(+x\) -direction) to stretch the spring, and released. What is the speed of the object when it is 0.50 \(\mathrm{m}\) to the right of the \(x=0\) equilibrium position?

You are designing a delivery ramp for crates containing exercise equipment. The \(1470-\mathrm{N}\) crates will move at 1.8 \(\mathrm{m} / \mathrm{s}\) at the top of a ramp that slopes downward at \(22.0^{\circ} .\) The ramp exerts a 550 -N kinetic friction force on each crate, and the maximum static friction force also has this value. Each crate will compress a spring at the bottom of the ramp and will come to rest after traveling a total distance of 8.0 m along the ramp. Once stopped, a crate must not rebound back up the ramp. Calculate the force constant of the spring that will be needed in order to meet the design criteria.

Down the Pole. A fireman of mass \(m\) slides a distance \(d\) down a pole. He sturts from rest. He moves as fast at the bottom as if he had stepped off a platform a distance \(h \leq d\) above the ground and descended with negligible air resistance. (a) What average friction force did the fireman exert on the pole? Does your answer make sense in the special cases of \(h=d\) and \(h=0 ?\) (b) Find a numerical value for the average friction force a 75 -kg fireman exerts, for \(d=2.5 \mathrm{m}\) and \(h=1.0 \mathrm{m}\) (c) In terms of \(g, h,\) and \(d,\) what is the speed of the fireman when he is a distance \(y\) above the bottom of the pole?

CP A 0.300 -kg potato is tied to a string with length \(2.50 \mathrm{m},\) and the other end of the string is tied to a rigid support. The potato is held straight out horizontally from the point of support, with the string pulled taut, and is then released. (a) What is the speed of the potato at the lowest point of its motion? (b) What is the tension in the string at this point?

A spring stores potential energy \(U_{0}\) when it is compressed a distance \(x_{0}\) from its uncompressed length. (a) In terms of \(U_{0},\) how much energy does it store when it is compressed (i) twice as much and (ii) half as much? (b) In terms of \(x_{0},\) how much must it be compressed from its uncompressed length to store (i) twice as much energy and (ii) half as much energy?
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