/*! 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 15 Figure 9-44 shows an arrangement... [FREE SOLUTION] | 91Ó°ÊÓ

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Figure 9-44 shows an arrangement with an air track, in which a cart is connected by a cord to a hanging block. The cart has mass \(m_{1}=0.600 \mathrm{~kg}\), and its center is initially at \(x y\) coordinates \((-0.500\) \(\mathrm{m}, 0 \mathrm{~m}) ;\) the block has mass \(m_{2}=0.400 \mathrm{~kg}\), and its center is initially at \(x y\) coordinates \((0,-0.100 \mathrm{~m})\). The mass of the cord and pulley are negligible. The cart is released from rest, and both cart and block move until the cart hits the pulley. The friction between the cart and the air track and between the pulley and its axle is negligible. (a) In unit-vector notation, what is the acceleration of the center of mass of the cart-block system? (b) What is the velocity of the com as a function of time \(t ?\) (c) Sketch the path taken by the com. (d) If the path is curved, determine whether it bulges upward to the right or downward to the left, and if it is straight, find the angle between it and the \(x\) axis.

Short Answer

Expert verified
The acceleration of com is \((3.92 \hat{i}, -3.92 \hat{j})\) m/s²; it moves in a straight line at 45° to the x-axis.

Step by step solution

01

Understand the Problem

The problem involves a system of two masses connected by a cord running over a frictionless pulley. The cart (mass \(m_1 = 0.600 \text{ kg}\)) moves horizontally, and the block (mass \(m_2 = 0.400 \text{ kg}\)) moves vertically due to gravity. We need to find the center of mass acceleration, its trajectory, and conclude if the path is straight or curved.
02

Calculate System Acceleration

Both objects accelerate at the same rate because they are connected. The net force causing the acceleration is the gravitational force on the block: \(F = m_2g\), where \(g = 9.8 \text{ m/s}^2\). Thus, the acceleration \(a\) of the system can be determined by \[a = \frac{m_2g}{m_1+m_2} = \frac{0.400 \times 9.8}{0.600 + 0.400} = 3.92 \text{ m/s}^2.\] Since the system is symmetrical, the center of mass acceleration in vector notation is \( \mathbf{a}_{\text{com}} = (3.92 \hat{i}, -3.92 \hat{j}) \text{ m/s}^2 \).
03

Velocity of the Center of Mass

The velocity of the center of mass is obtained by integrating the acceleration with respect to time.\[\mathbf{v}_{\text{com}}(t) = \int \mathbf{a}_{\text{com}} \, dt = (3.92t \hat{i}, -3.92t \hat{j}),\] assuming the initial velocity of the center of mass is zero.
04

Path of the Center of Mass

The path in the \(xy\)-plane of the center of mass can be found by considering the velocity components. Because \(v_{x} = 3.92t\) and \(v_y = -3.92t\), it means the path equation is linear, \[y = -x.\]
05

Nature of the Path

Since the path equation \(y = -x\) is a straight line, it confirms that the path taken by the center of mass is straight. The negative sign indicates it is decreasing and the angle with the \(x\)-axis is 45 degrees downward, to the left.

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Key Concepts

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

Newton's Second Law
Isaac Newton's Second Law of Motion is a fundamental principle in physics. It tells us how the velocity of an object changes when it is subjected to external forces. According to this law, the acceleration (\( \mathbf{a} \)) of an object is directly proportional to the net force (\( \mathbf{F} \)) acting on it and inversely proportional to its mass (\( m \)). Mathematically, this is expressed as:\[ \mathbf{F} = m \mathbf{a} \].
In the given problem, a cart and a block are connected via a pulley. The gravitational force pulling on the block acts as the net external force.
By Newton's Second Law, this force causes the system to accelerate. Understanding this connection between force and motion can help you predict how objects behave when forces are applied. When you think about why a car speeds up as you press the gas pedal, it's because the engine generates force, causing acceleration as per Newton's principle.
In this exercise, both the cart and the block move together, demonstrating how interconnected bodies accelerate together because of a shared force.
Linear Motion
Linear motion occurs when an object moves along a straight path. In this scenario, the center of mass of the cart-block system exhibits linear motion. This is evident from the solution that states the path is a straight line \[ y = -x \].
Linear motion can be described using basic equations of motion. When an object moves at constant acceleration, its position, velocity, and time are intertwined by these relationships.
For this exercise, since both objects start from rest, their initial velocity is zero. From the problem's solution, once you calculate the constant acceleration using Newton's Second Law, you can then find how velocity changes over time.
Linear motion can be felt in everyday life, such as when you drive down a highway or run a straight track. Understanding it helps to make predictions and solve real-world physics problems.
Gravitational Force
Gravitational force is a force of attraction between two masses. Here, it plays a crucial role in pulling the block downward, thereby enabling the entire system to move. The force is given by \( F = m_2 g \), where \( g \) is the acceleration due to Earth's gravity, approximated as \( 9.8 \text{ m/s}^2 \).
In this problem, the cart and block are part of a system where gravity primarily acts on the block, making it fall downward. This force, transferred through the cord and pulley, causes the cart to move horizontally.
Gravitational force is not only essential in this exercise but is also a fundamental force in the universe. It keeps planets in orbit around the Sun and governs the motion of celestial bodies. In labs, this force can be measured to understand the behavior of different objects and materials under Earth’s gravity. Knowing how it works is essential for fields like engineering, astronomy, and physics.

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Most popular questions from this chapter

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