/*! 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 151 A horizontal force of \(10 \math... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 151

A horizontal force of \(10 \mathrm{~N}\) is necessary to just hold a block stationary against a wall. The co-efficient of friction between the block and wall is \(0.2\). The weight of the block is (A) \(20 \mathrm{~N}\) (B) \(50 \mathrm{~N}\) (C) \(100 \mathrm{~N}\) (D) \(2 \mathrm{~N}\)

Short Answer

Expert verified
The weight of the block is \(2 \mathrm{~N}\) (Option D).

Step by step solution

01

Identify the forces acting on the block

Since the block is stationary against the wall, there are three forces acting on it: 1. The gravitational force (weight) acting downward. 2. The horizontal force (provided in the problem) holding the block against the wall. 3. The frictional force acting upward, preventing the block from sliding down the wall.
02

Write down the equations for frictional force and normal force

The frictional force (F_f) is given by the equation: \(F_f = \mu F_n\), where \(\mu\) is the coefficient of friction and \(F_n\) is the normal force. In this problem, the normal force is equal to the applied horizontal force (F_H), since it opposes the horizontal force: \(F_n = F_H\).
03

Determine the weight of the block using the given information

We are given that the horizontal force necessary to hold the block stationary against the wall is \(10 \mathrm{~N}\), and the coefficient of friction is \(0.2\). We can use these values to determine the weight (W) of the block. First, let's find the normal force: \(F_n = F_H = 10 \mathrm{~N}\). Now, we can find the frictional force acting on the block using the formula: \(F_f = \mu F_n = 0.2 * 10 \mathrm{~N} = 2 \mathrm{~N}\). Since the block is stationary against the wall, the frictional force is equal to the weight of the block: \(W = F_f = 2 \mathrm{~N}\). Therefore, the weight of the block is 2 N (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!

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 angular velocity of a wheel increases from 1200 rpm to 4500 rpm in \(10 \mathrm{~s}\). The number of revolutions made during this time is (A) 950 (B) 475 (C) \(237.5\) (D) \(118.75\)

A particle is moving along the circular path with a speed \(v\) and tangential acceleration is \(g\) at an instant. If the radius of the circular path be \(r\), then the net acceleration of the particle at that instant is (A) \(\frac{v^{2}}{r}+g\) (B) \(\frac{v^{2}}{r^{2}}+g^{2}\) (C) \(\left[\frac{v^{4}}{r^{2}}+g^{2}\right]^{\frac{1}{2}}\) (D) \(\left[\frac{v^{2}}{r}+g^{2}\right]^{\frac{1}{2}}\)

A particle is moving with a constant angular acceleration of \(4 \mathrm{rad} / \mathrm{s}^{2}\) in a circular path. At \(t=0\), particle was at rest. Find the time at which the magnitudes of centripetal acceleration and tangential acceleration are equal. \(\begin{array}{llll}\text { (A) } 1 \mathrm{~s} & \text { (B) } 2 \mathrm{~s} & \text { (C) } \frac{1}{2} \mathrm{~s} & \text { (D) } \frac{1}{4} \mathrm{~s}\end{array}\)

Two masses \(m\) and \(M\) are connected by a light string passing over a smooth pulley. When set free, \(m\) moves up by \(1.4 \mathrm{~m}\) in \(2 \mathrm{~s}\). The ratio \(\frac{m}{M}\) is \(\left(g=9.8 \mathrm{~ms}^{-2}\right.\) ) (A) \(\frac{13}{15}\) (B) \(\frac{15}{13}\) (C) \(\frac{9}{7}\) (D) \(\frac{7}{9}\)

A body of mass \(5 \mathrm{~kg}\) is acted upon by two perpendicular forces \(8 \mathrm{~N}\) and \(6 \mathrm{~N}\). Give the magnitude of the acceleration of the body. (A) \(2 \mathrm{~m} / \mathrm{s}^{2}\) (B) \(4 \mathrm{~m} / \mathrm{s}^{2}\) (C) \(6 \mathrm{~m} / \mathrm{s}^{2}\) (D) \(8 \mathrm{~m} / \mathrm{s}^{2}\)
See all solutions

Recommended explanations on Physics Textbooks

Magnetism and Electromagnetic Induction

Read Explanation

Force

Read Explanation

Linear Momentum

Read Explanation

Space Physics

Read Explanation

Electrostatics

Read Explanation

Famous Physicists

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.