/*! 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 83 Two blocks are suspended from op... [FREE SOLUTION] | 91Ó°ÊÓ

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Two blocks are suspended from opposite ends of a light rope that passes over a light, friction less pulley. One block has mass \(m_{1}\) and the other has mass \(m_{2},\) where \(m_{2}>m_{1}\). The two blocks are released from rest, and the block with mass \(m_{2}\) moves downward \(5.00 \mathrm{~m}\) in \(2.00 \mathrm{~s}\) after being released. While the blocks are moving, the tension in the rope is \(16.0 \mathrm{~N}\). Calculate \(m_{1}\) and \(m_{2}\).

Short Answer

Expert verified
The two masses \(m_{1}\) and \(m_{2}\) can be computed by solving the system of equations from Newton's second law applied to each mass.

Step by step solution

01

Calculate the acceleration of the system

The acceleration \(a\) of the system can be determined by using the second equation of motion \(d = ut + 0.5at^2\). Here, distance \(d = 5.00 m,\) time \(t = 2.00 s,\) and initial velocity \(u = 0\) as it is released from rest. Therefore, the acceleration \(a\) is calculated as \(a = 2d / t^2\).
02

Apply Newton's second law on \(m_{2}\)

Newton's second law states that the net force on an object is equal to its mass times its acceleration \(F=ma\). Hence, on \(m_{2}\), force due to \(m_{2}\) is \(m_{2}g\) downwards and tension \(T\) upwards. As \(m_{2}\) is moving downwards, the net force \(F\) on \(m_{2}\) would be \(m_{2}g - T\). But this is equal to \(m_{2}a\). Hence, \(m_{2} = (T + m_{2}a) / g\).
03

Apply Newton's second law on \(m_{1}\)

Similarly, applying Newton's second law on mass \(m_{1}\), the force due to \(m_{1}\) is \(m_{1}g\) downwards and tension \(T\) upwards. As \(m_{1}\) is moving upwards, the net force \(F\) on \(m_{1}\) would be \(T - m_{1}g\). But this is equal to \(m_{1}a\). Hence, \(m_{1} = (T - m_{1}a) / g\).
04

Solve the system of equations

Now, we have two equations of \(m_{1}\) and \(m_{2}\). Solve these two equations simultaneously to get the values of \(m_{1}\) and \(m_{2}\).

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Key Concepts

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

Acceleration Calculation
When we talk about acceleration in physics, we're discussing how quickly something speeds up or slows down. It’s a key concept when analyzing motion and is critical in a range of mechanics problems.
To determine acceleration, you would often use the formula given by the second equation of motion:
  • \[ d = ut + \frac{1}{2} a t^2 \]

In the case of the problem with two blocks and a pulley, the blocks start from rest meaning the initial velocity, \(u\), is zero. So, the formula simplifies to \(d = \frac{1}{2} a t^2\).
Knowing the distance \(d = 5.00 \text{ m}\) and time \(t = 2.00 \text{ s}\), you can solve for acceleration \(a\) using:
  • \[ a = \frac{2d}{t^2} \]
This equation helps us understand how the blocks speed up as they move along the rope. Calculating the acceleration is the first step in analyzing their motion.
Tension in Ropes
Tension is a force exerted by a rope when it is used to transmit force. In mechanics problems involving pulleys, understanding tension is crucial as it affects the movement of objects connected by the rope. In our problem, the tension in the rope is given as 16.0 N.
Consider block \(m_{2}\), which moves downward. The gravitational force acting on it is \(m_{2}g\), while tension in the rope works against gravity. Since \(m_{2}\) accelerates downward, the net force (tension subtracted from gravity) is equal to the mass of the block times its acceleration \(m_{2}a\).
This can be formulated as:
  • \[ m_{2}g - T = m_{2}a \]
On the other hand, block \(m_{1}\), which moves upward, is affected by tension pulling it up against gravity. Again, the net force is the difference between tension and gravitational force, translated to the block's upward acceleration:
  • \[ T - m_{1}g = m_{1}a \]
These equations show how tension helps balance the forces in the system, allowing us to solve for the masses involved.
Mechanics Problems
Mechanics problems like this one are classic examples in physics that help illustrate Newton's Laws of Motion. They often involve careful examination of forces and motions to solve complex systems. In our problem, we use Newton’s Second Law \(F = ma\) as a starting point.
Let’s break down a mechanics problem involving two masses connected by a rope over a pulley:
  • Identify forces: Each block has gravitational force (weight) and tension acting on it. For block \(m_{2}\), these forces are in opposite directions.
  • Apply Newton’s laws: Establish equations for each block using \(F = ma\).
  • Solve equations: Use the two equations derived from the forces acting on each block to find unknowns like mass or acceleration. This often involves substituting one equation into the other to find a common variable.
Understanding and solving mechanics problems require a systematic approach, where identifying all the forces and using physics laws guide you to a solution.

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Most popular questions from this chapter

Jack sits in the chair of a Ferris wheel that is rotating at a constant \(0.100 \mathrm{rev} / \mathrm{s}\). As Jack passes through the highest point of his circular path, the upward force that the chair exerts on him is equal to one- fourth of his weight. What is the radius of the circle in which Jack travels? Treat him as a point mass.

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A box of mass \(12.0 \mathrm{~kg}\) sits at rest on a horizontal surface. The coefficient of kinetic friction between the surface and the box is \(0.300 .\) The box is initially at rest, and then a constant force of magnitude \(F\) and direction \(37.0^{\circ}\) below the horizontal is applied to the box; the box slides along the surface. (a) What is \(F\) if the box has a speed of \(6.00 \mathrm{~m} / \mathrm{s}\) after traveling a distance of \(8.00 \mathrm{~m} ?\) (b) What is \(F\) if the surface is friction less and all the other quantities are the same? (c) What is \(F\) if all the quantities are the same as in part (a) but the force applied to the box is horizontal?

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