/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 Two blocks of masses \(6 \mathrm... [FREE SOLUTION] | 91Ó°ÊÓ

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Two blocks of masses \(6 \mathrm{~kg}\) and \(4 \mathrm{~kg}\) are attached to the two ends of a massless string passing over a smooth fixed pulley. If the system is released, the acceleration of the centre of mass of the system will be (1) \(g\), vertically downwards (2) \(g / 5\), vertically downwards (3) \(g / 25\), vertically downwards (4) zero

Short Answer

Expert verified
The acceleration of the center of mass is \(g/5\), vertically downwards.

Step by step solution

01

Calculate Net Force Acting on the System

Identify the forces acting on the blocks. The tension in the string is the same on each side but acts in opposite directions. Each block has a gravitational force acting on it, which is its weight: \( W_1 = m_1 \cdot g \) for the 6 kg block and \( W_2 = m_2 \cdot g \) for the 4 kg block. The net force \( F_{net} \) acting on the system is \( F_{net} = m_1 \cdot g - m_2 \cdot g \).
02

Compute Total Mass of the System

Calculate the total mass of the two-block system: \( m_{total} = m_1 + m_2 = 6 \text{ kg} + 4 \text{ kg} = 10 \text{ kg} \).
03

Calculate System Acceleration

Use Newton's second law to find the acceleration of the system: \( F_{net} = m_{total} \cdot a \). From Step 1, \( F_{net} = (m_1 - m_2) \cdot g = (6 \text{ kg} - 4 \text{ kg}) \cdot g = 2 \cdot g \). Now apply \( a = \frac{F_{net}}{m_{total}} = \frac{2 \cdot g}{10} = \frac{g}{5} \).
04

Determine Acceleration of Center of Mass

The center of mass of the system will accelerate at the same rate as the system's acceleration because it's a simple two-body system on a frictionless pulley. Therefore, the acceleration of the center of mass is \( \frac{g}{5} \), vertically downwards.

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Key Concepts

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

Newton's Second Law
Newton's Second Law forms the foundation for understanding motion in a pulley system. Simply put, this law states that the acceleration of an object is dependent on the net force acting upon it and its mass. Its mathematical expression is \[ F = m \cdot a \] where
  • \( F \) is the net force,
  • \( m \) is the mass of the object,
  • and \( a \) represents the acceleration.
Applying this law to our pulley system, when the two blocks are suspended on either side, the larger force from the heavier block will cause movement.
The principle is crucial here as it orchestrates how net force translates into motion across the pulley, dictating the velocity and direction of the blocks' movement. By using this law, we can predict how forces influence acceleration in real-world scenarios with precision.
Center of Mass Acceleration
The center of mass of a system is a point where all its mass is concentrated and behaves as if it were a single mass. In the context of our pulley system with two blocks, understanding the center of mass acceleration is about envisioning how the entire system moves.
The important fact to note is that the center of mass itself will move at the same rate as the system's calculated acceleration. This happens because each component's motion is symmetrically accounted for, meaning the net external force acting on the entire system drives its center of mass uniformly.
The acceleration of the center of mass simplifies calculations of motion for complex systems. By applying Newton's second law to our system, we derive the acceleration it experiences.
For this pulley setup, it ultimately emerges as \[ a_{cm} = \frac{F_{net}}{m_{total}} = \frac{g}{5} \]. This computation shows how the combined mass affects its central movement, illustrating overall dynamics efficiently.
Net Force Calculation
Calculating net force is an integral part of determining system motion dynamics. In a pulley system, net force considers all forces acting on the masses involved. For our scenario, the forces in play are due to gravity acting on each block.
The expression \[ F_{net} = m_1 \cdot g - m_2 \cdot g \]helps us understand this, where
  • \( m_1 \) and \( m_2 \) are the masses (6 kg and 4 kg respectively),
  • and \( g \) represents gravitational acceleration.
Calculating it provides \[ F_{net} = (6 - 4) \cdot g = 2 \cdot g \]. Net force results from the imbalance in gravitational pull between both blocks that causes acceleration.
This imbalance is what drives the system's motion and allows us to predict its behavior using Newton's laws. Understanding these dynamics is essential, as it influences the acceleration and eventually how the system behaves overall.

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Most popular questions from this chapter

A body of mass \(1 \mathrm{~kg}\) initially at rest, explodes and breaks into three fragments of masses in the ratio \(1: 1: 3\). The two pieces of equal mass fly off perpendicular to each other with a speed of \(15 \mathrm{~ms}^{-1}\) each. What is the velocity of the heavier fragment? (1) \(10 \sqrt{2} \mathrm{~m} \mathrm{~s}^{-1}\) (2) \(5 \sqrt{3} \mathrm{~ms}^{-1}\) (3) \(10 \sqrt{3} \mathrm{~ms}^{-1}\) (4) \(5 \sqrt{2} \mathrm{~ms}^{-1}\)

Two blocks \(m_{1}\) and \(m_{2}\) are pulled on a smooth horizontal surface, and are joined together with a spring of stiffness \(k\) as shown in figure. Suddenly, block \(m_{2}\) receives a horizontal velocity \(v_{0}\), then the maximum extension \(x_{m}\) in the spring is (1) \(v_{0} \sqrt{\frac{m_{1} m_{2}}{m_{1}+m_{2}}}\) (2) \(v_{0} \sqrt{\frac{2 m_{1} m_{2}}{\left(m_{1}+m_{2}\right) k}}\) (3) \(v_{0} \sqrt{\frac{m_{1} m_{2}}{2\left(m_{1}+m_{2}\right) k}}\) (4) \(v_{0} \sqrt{\frac{m_{1} m_{2}}{\left(m_{1}+m_{2}\right) k}}\)

An object comprises of a uniform ring of radius \(R\) and its uniform chord \(A B\) (not necessarily made of the same material) as shown. Which of the following can not be the centre of mass of the object? (1) \(\left(\frac{R}{3}, \frac{R}{3}\right)\) (2) \(\left(\frac{R}{3}, \frac{R}{2}\right)\) (3) \(\left(\frac{R}{4}, \frac{R}{4}\right)\) (4) None of these

Two men ' \(A\) ' and ' \(B\) ' of mass \(50 \mathrm{~kg}\) and \(70 \mathrm{~kg}\) respectively are standing on the ends of a plank of mass \(80 \mathrm{~kg}\). The length of plank is \(5 \mathrm{~m}\) and it is kept on a smooth horizontal surface. Now man starts moving and exchange their positions on the plank. Then (1) The distance moved by centre of mass of the system ' \(A\) ' \(+' B^{\prime}+\) plank is \(50 \mathrm{~cm}\) (2) The displacement of the plank is \(50 \mathrm{~cm}\) (toward right) (3) The distance moved by man ' \(A\) ' with respect to ground is \(5.5 \mathrm{~m}\) (4) The distance moved by man ' \(B\) ' with respect to ground is \(6 \mathrm{~m}\).

Consider a hollow sphere of mass \(M\) and radius \(R\) resting on a smooth surface. A smaller sphere of mass \(m\) and radius \(t\) is initially held at position \(P\) within the bigger sphere. If the small sphere is now released, it rolls down the inner surface of hollow sphere and finally stops at its bottom point \(Q\). Then: (1) horizontal displacement of the smaller sphere on smooth surface is \(\frac{M(R-r)}{M+m}\) (2) horizontal displacement of the bigger sphere on smooth surface is \(\frac{M(R+r)}{M+m}\) (3) horizontal displacement of the bigger sphere on smooth surface is \(\frac{m(R-r)}{M+m}\) (4) none of these

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