/*! 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 44 A block of mass \(m\) is pulled ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 44

A block of mass \(m\) is pulled by a constant power \(P\) placed on a rough horizontal plane. The friction co-efficient between the block and surface varies with its speed \(v\) as \(\mu=\frac{1}{\sqrt{1+v}} .\) The acceleration of the block when its speed is \(3 \mathrm{~m} / \mathrm{s}\) will be (A) \(\frac{P}{3 m}-\frac{g}{2}\) (B) \(\frac{P}{3 m}+\frac{g}{2}\) (C) \(\frac{P}{3 m}\) (D) \(\frac{g}{2}\)

Short Answer

Expert verified
The short answer to the problem is: \(a = \frac{P}{3m} - \frac{g}{2}\).

Step by step solution

01

Write down the expression for power and friction force

The power P is given by the product of force F and speed v: \(P = Fv\) The friction force f is given by 渭 times the normal force (which is equal to the block's weight W), where \(W = mg\), and \(f = \mu W = \mu mg\)
02

Write down Newton's second law of motion for the block

Newton's second law states that the net force acting on an object is equal to its mass times its acceleration. In this case, the net force is the difference between the applied force F and the friction force f: \(F - f = ma\)
03

Find the expression for acceleration in terms of given parameters

First, we replace friction force f in the equation from Step 2: \(F - \mu mg = ma\) Next, we replace the force F using the expression from Step 1: \(P/v - \mu mg = ma\) Now, we substitute the given expression for 渭: \(\frac{P}{v} - \frac{mg}{\sqrt{1+v}} = ma\)
04

Calculate the acceleration when v = 3 m/s

We are given the speed v = 3 m/s. Substitute this value in the equation from Step 3: \(\frac{P}{3} - \frac{mg}{\sqrt{1+3}} = ma\) Simplify the equation: \(\frac{P}{3} - \frac{mg}{2} = ma\) Now divide both sides by m to get the acceleration a: \(a = \frac{P}{3m} - \frac{g}{2}\) So, the acceleration of the block when its speed is 3 m/s is equal to \(\frac{P}{3m} - \frac{g}{2}\). Therefore, the correct answer is (A).

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!

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

The co-efficient of friction between the block and plank is \(\mu\) and its value is such that block becomes stationary with respect to plank before it reaches the other end. Then (A) the work done by friction on the block is negative. (B) the work done by friction on the plank is positive. (C) the net work done by friction is negative. (D) net work done by the friction is zero.

A \(2 \mathrm{~kg}\) block slides on a horizontal floor with a speed of \(4 \mathrm{~m} / \mathrm{s}\). It strikes an uncompressed spring, and compresses it till the block is motionless. The kinetic friction force is \(15 \mathrm{~N}\) and spring constant is \(10,000 \mathrm{~N} / \mathrm{m}\). The spring compresses by \([\mathbf{2 0 0 7}]\) (A) \(8.5 \mathrm{~cm}\) (B) \(5.5 \mathrm{~cm}\) (C) \(2.5 \mathrm{~cm}\) (D) \(11.0 \mathrm{~cm}\)

Two springs \(A\) and \(B\left(k_{A}=2 k_{B}\right)\) are stretched by applying forces of equal magnitudes at the ends. If the energy stored in \(A\) is \(E\), then energy stored in \(B\) is (A) \(\frac{E}{2}\) (B) \(2 E\) (C) \(E\) (D) \(\frac{E}{4}\)

The block of mass \(m\) moving on the frictionless horizontal surface collides with the spring of spring constant \(k\) and compresses it by length \(L\). The maximum momentum of the block after collision is \([\mathbf{2 0 0 5}]\) (A) \(\frac{k L^{2}}{2 m}\) (B) \(\sqrt{m k} L\) (C) \(\frac{m L^{2}}{k}\) (D) Zero

A block of mass \(m\) is pulled by a constant power \(P\) placed on a rough horizontal plane. The friction co-efficient between the block and surface is \(\mu\). The maximum velocity of the block is (A) \(\frac{P}{m g}\) (B) \(\frac{P}{\mu m g}\) (C) \(\frac{\mu P}{m g}\) (D) Infinite
See all solutions

Recommended explanations on Physics Textbooks

Oscillations

Read Explanation

Thermodynamics

Read Explanation

Famous Physicists

Read Explanation

Modern Physics

Read Explanation

Engineering Physics

Read Explanation

Solid State Physics

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.