Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 352

Two masses \(\mathrm{M}\) and \((\mathrm{M} / 2)\) are joined together by means of light inextensible string passed over a frictionless pulley as shown in fig. When the bigger mass is released, the small one will ascend with an acceleration (A) \((\mathrm{g} / 3)\) (B) \((3 \mathrm{~g} / 2)\) (C) \(\mathrm{g}\) (D) \((\mathrm{g} / 2)\)

Short Answer

Expert verified
The acceleration of the smaller mass is \(a = \frac{g}{3}\), so the correct answer is (A) \(\frac{g}{3}\).

Step by step solution

01

Define the forces acting on each mass

First, let's identify and label the forces acting on each mass. For mass M, there is a downward gravitational force, which can be expressed as \(F_{Mg} = M \cdot g\), where g is the acceleration due to gravity. Similarly, for mass M/2, there is a downward gravitational force \(F_{(M/2)g} = (M/2) \cdot g\). There is also tension in the string, which we will denote as T. The tension acts upward on both masses.
02

Write the equations of motion for each mass

According to Newton's second law of motion, the net force acting on an object is equal to its mass multiplied by its acceleration. Therefore, we can write the equations of motion for both masses: 1. For mass M: \(M \cdot a = M \cdot g - T\) 2. For mass M/2: \((M/2) \cdot a = T - (M/2) \cdot g\) Now we have a system of two equations with two unknowns (acceleration a and tension T).
03

Solve the system of equations

We will first solve the equation for mass M to express the tension T in terms of acceleration a: \(T = M \cdot g - M \cdot a\) Now, substitute this expression for tension T in the equation for mass M/2: \((M/2) \cdot a = (M \cdot g - M \cdot a) - (M/2) \cdot g\)
04

Solve for acceleration a

Now, we can solve the equation for acceleration a: \((M/2) \cdot a + M \cdot a = M \cdot g - (M/2) \cdot g\) \((3/2)Ma = (1/2)Mg\) Divide both sides by \(3/2M\): \(a = \frac{1/2 \cdot g}{3/2} = \frac{g}{3}\) The acceleration of the smaller mass is \(a = \frac{g}{3}\). So the correct answer is (A) \(\frac{g}{3}\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 block of mass \(4 \mathrm{~kg}\) is placed on a rough horizontal plane. A time dependent force \(\mathrm{F}=\mathrm{Kt}^{2}\) acts on a block, where \(\mathrm{k}=2 \mathrm{~N} / \mathrm{s}^{2}\), co-efficient of friction \(\mu=0.8\). Force of friction between the block and the plane at $\mathrm{t}=2 \mathrm{~S}$ is.... (A) \(32 \mathrm{~N}\) (B) \(4 \mathrm{~N}\) (C) \(2 \mathrm{~N}\) (D) \(8 \mathrm{~N}\)

A vehicle of \(100 \mathrm{~kg}\) is moving with a velocity of $5(\mathrm{~m} / \mathrm{s})\(. To stop it in \)(1 / 10) \mathrm{sec}$, the required force in opposite direction is \(\mathrm{N}\) (A) 50 (B) 500 (C) 5000 (D) 1000

A wagon weighing \(1000 \mathrm{~kg}\) is moving with a velocity $50 \mathrm{~km} \mathrm{~h}^{-1}\( on smooth horizontal rails. A mass of \)250 \mathrm{~kg}$ is dropped into it. The velocity with which it moves now is (A) \(2.5 \mathrm{~km} \mathrm{~h}^{-1}\) (B) \(20 \mathrm{~km} \mathrm{~h}^{-1}\) (C) \(40 \mathrm{~km} \mathrm{~h}^{-1}\) (D) \(50 \mathrm{~km} \mathrm{~h}^{-1}\)

A force of \(8 \mathrm{~N}\) acts on an object of mass \(5 \mathrm{~kg}\) in \(\mathrm{X}\) -direction and another force of \(6 \mathrm{~N}\) acts on it in \(\mathrm{Y}\) -direction. Hence, the magnitude of acceleration of object will be (A) \(1.5 \mathrm{~ms}^{-2}\) (B) \(2.0 \mathrm{~ms}^{-2}\) (C) \(2.5 \mathrm{~ms}^{-2}\) (D) \(3.5 \mathrm{~ms}^{-2}\)

A ball falls on surface from \(10 \mathrm{~m}\) height and rebounds to $2.5 \mathrm{~m} .\( If duration of contact with floor is \)0.01 \mathrm{sec}$. then average acceleration during contact is \(\mathrm{ms}^{-2}\) (A) 2100 (B) 1400 \(\begin{array}{ll}\text { (C) } 700 & \text { (D) } 400\end{array}\)
See all solutions

Recommended explanations on English Textbooks

Lexis and Semantics Summary

Read Explanation

Rhetoric

Read Explanation

Morphology

Read Explanation

5 Paragraph Essay

Read Explanation

Language and Social Groups

Read Explanation

Semiotics

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.

Sign-up for free