/*! 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 120 The masses of the blocks \(A\) a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 120

The masses of the blocks \(A\) and \(B\) are \(m\) and \(M .\) Between \(A\) and \(B\) there is a constant frictional force \(F\), and \(B\) can slide frictionlessly on horizontal surface (see Fig. \(7.378\) ). \(A\) is set in motion with velocity while \(B\) is at rest. What is the distance moved by \(A\) relative to \(B\) before they move with the same velocity? a. \(\frac{m M v_{0}^{2}}{F(m-M)}\) b. \(\frac{m M v_{0}^{2}}{2 F(m-M)}\) c. \(\frac{m M v_{0}^{2}}{F(m+M)}\) d. \(\frac{m M v_{0}^{2}}{2 F(m+M)}\)

Short Answer

Expert verified
d. \(\frac{m M v_{0}^{2}}{2 F(m+M)}\)

Step by step solution

01

Understanding the Problem

The blocks A and B have masses \(m\) and \(M\) respectively. Block A has an initial velocity of \(v_0\), while block B is initially at rest. There is a frictional force \(F\) acting between these blocks, and block B can slide without friction on the surface. We need to find the distance moved by block A relative to block B before they both move at the same velocity.
02

Applying Conservation of Momentum

As there are no external horizontal forces, the total momentum of the system (A + B) is conserved. Initially, only block A has momentum, \(mv_0\). When blocks A and B have the same final velocity \(v_f\), the total momentum is \((m+M)v_f\). Thus:\[ mv_0 = (m + M)v_f \] Solving for \(v_f\):\[ v_f = \frac{mv_0}{m+M} \]
03

Using Work-Energy Principle

The work done by the frictional force \(F\) brings the blocks to a common velocity. The initial kinetic energy of block A is \(\frac{1}{2}mv_0^2\) and the final kinetic energy of both blocks together is \(\frac{1}{2}(m+M)v_f^2\). Using the work-energy principle:\[ Fd = \frac{1}{2}mv_0^2 - \frac{1}{2}(m+M)v_f^2 \]Where \(d\) is the relative distance moved before having the same velocity.
04

Solving the Work-Energy Equation

Substitute \(v_f = \frac{mv_0}{m+M}\) into the energy equation:\[ Fd = \frac{1}{2}mv_0^2 - \frac{1}{2}(m+M)\left(\frac{mv_0}{m+M}\right)^2 \]Simplifying the terms:\[ = \frac{1}{2}mv_0^2 - \frac{1}{2} \frac{m^2 v_0^2}{m+M} \]\[ = \frac{1}{2}mv_0^2 (1 - \frac{m}{m+M}) \]\[ = \frac{1}{2}mv_0^2 \left(\frac{M}{m+M}\right) \]Substituting this back, we get:\[ Fd = \frac{1}{2} \frac{mMv_0^2}{m+M} \]
05

Solving for Distance

Rearranging the equation \(Fd = \frac{1}{2} \frac{mMv_0^2}{m+M} \), we solve for \(d\): \[ d = \frac{1}{2} \frac{mMv_0^2}{F(m+M)} \]Thus the distance moved by A relative to B before they move with the same velocity is option (d).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Conservation of Momentum
In physics, the Conservation of Momentum principle is a fundamental concept that helps us understand how objects behave when they interact. It states that the total momentum of an isolated system remains constant if no external forces are acting on it. Momentum is the product of the mass and velocity of an object, and it is a vector, meaning it has direction as well as magnitude.

When analyzing systems like the two blocks in the exercise, this principle shows that the initial momentum of the moving block must equal the total momentum of both blocks once they reach the same speed. It means:
  • Initial momentum = Final momentum
  • Before interaction, only A has momentum: \( mv_0 \)
  • After interaction, both A and B have the same velocity: \((m + M)v_f\)
This leads us to the equation \( mv_0 = (m + M)v_f \), which we use to understand how the two blocks share momentum as they interact.
Work-Energy Principle
The Work-Energy Principle connects the concepts of work and kinetic energy, showing how work done on an object changes its kinetic energy. Work is the process by which energy is transferred from one object or system to another, usually exerted by forces over a distance.

In the context of the exercise, we consider the frictional force doing work on the blocks to bring them to the same velocity. Initially, block A has kinetic energy \( \frac{1}{2}mv_0^2 \). When both blocks move together, their combined kinetic energy is \( \frac{1}{2}(m+M)v_f^2 \).
  • The work done by the frictional force \( F \) changes the system's kinetic energy.
  • Equation: \( Fd = \frac{1}{2}mv_0^2 - \frac{1}{2}(m+M)v_f^2 \)
By solving this equation, we can find the distance \( d \) that block A travels relative to block B, with the change in kinetic energy being directly linked to the work done by friction.
Frictional Force
Friction is a force that opposes the relative motion of two surfaces in contact. It is an essential component in this exercise because it acts between blocks A and B, influencing how they reach the same final speed.

Although block B slides without friction on the horizontal surface, the frictional force \( F \) between the two blocks is crucial. This constant force slows down block A and speeds up block B until they match velocities. Understanding friction helps in predicting how energy transfers and momentum changes:
  • It opposes the motion of block A.
  • It acts as a catalyst, allowing energy to be transferred from block A to block B.
  • It is the key to completing the energy balance equation in our work-energy analysis.
Friction is often seen as an opposing force in mechanics, but it plays a crucial role in enabling systems to reach equilibrium, as demonstrated by the motion of these blocks.
Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. It is given by the formula \( \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is the velocity of the object. In this exercise, kinetic energy is a central theme, helping us understand how energy transfers between the blocks as they interact.

Initially, only block A has kinetic energy due to its velocity \( v_0 \). As the blocks interact via the frictional force \( F \), kinetic energy is transferred:
  • Block A starts with energy \( \frac{1}{2}mv_0^2 \).
  • As energy transfers, block B starts moving, and eventually, both share total kinetic energy \( \frac{1}{2}(m+M)v_f^2 \).
This understanding of kinetic energy exchange is necessary to apply the work-energy principle and calculate the relative distance moved by block A. Through kinetic energy, we can quantify and better understand the movement and interaction of objects in physics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A body of mass \(m\) is launched up on a rough inclined plane making an angle \(45^{\circ}\) with horizontal. If the time of ascent is half of the time of descent, the frictional coefficient between plane and body is a. \(\frac{2}{5}\) b. \(\frac{3}{5}\) c. \(\frac{3}{4}\) d. \(\frac{4}{5}\)

A man sits on a chair supported by a rope passing over a frictionless fixed pulley. The man who weighs \(1.000 \mathrm{~N}\) exerts a force of \(450 \mathrm{~N}\) on the chair downwards while puiling the rope on the other side. If the chair weighs \(250 \mathrm{~N}\), then the acceleration of the chair is a. \(0.45 \mathrm{~m} / \mathrm{s}^{2}\) b. 0 c. \(2 \mathrm{~m} / \mathrm{s}^{2}\) d. \(9 / 25 \mathrm{~m} / \mathrm{s}^{2}\)

A pendulum of mass \(m\) hangs from a support fixed to a trolley. The direction of the string when the trolley rolls up a plane of inclination \(\alpha\) with acceleration \(a_{0}\) is a. \(\theta=\tan ^{-1} \alpha\) b. \(\theta=\tan ^{-1}\left(\frac{a_{0}}{g}\right)\) c. \(\theta=\tan ^{-1}\left(\frac{g}{a_{0}}\right)\) d. \(\theta=\tan ^{-1}\left(\frac{a_{0}+g \sin \alpha}{g \cos \alpha}\right)\)

A lift is moving down with an acceleration \(a .\) A man in the lift drops a ball inside the lift. The acceleration of the ball as observed by the man in the lift, and a man standing stationary on the ground are, respectively a. \(a, g\) b. \((g-a) ; g\) c. \(a_{1} a\) d. \(g, g\)

Three light strings are connected at the point \(P .\) A weight \(W\) is suspended from one of the strings. End \(A\) of string AP and end \(B\) of string \(P B\) are fixed as shown. In equilibrium \(P B\) is horizontal and \(P A\) makes an angle of \(60^{\circ}\) with the horizontal. If the tension in \(P B\) is \(30 \mathrm{~N}\) then the tension in \(P A\) and weight \(W\) are respectively given by a. \(60 \mathrm{~N} ; 30 \mathrm{~N}\) b. \(60 / \sqrt{3} \mathrm{~N} ; 30 / \sqrt{3} \mathrm{~N}\) c. \(60 \mathrm{~N} ; 30 \sqrt{3} \mathrm{~N}\) d. \(60 \sqrt{3} \mathrm{~N}: 30 \sqrt{3} \mathrm{~N}\)
See all solutions

Recommended explanations on Physics Textbooks

Electrostatics

Read Explanation

Turning Points in Physics

Read Explanation

Medical Physics

Read Explanation

Electromagnetism

Read Explanation

Radiation

Read Explanation

Atoms and Radioactivity

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.