/*! 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} Q.46 A long, 1.0聽kg聽rope hangs from... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Q.46 (page 180)

A long, 1.0kgrope hangs from a support that breaks, causing the rope to fall, if the pull exceeds 40N. A student team has built a 2.0kgrobot 鈥渕ouse鈥� that runs up and down the rope. What maximum acceleration can the robot have鈥攂oth magnitude and direction鈥攚ithout the rope falling?

Short Answer

Expert verified

The maximum acceleration can the robot have is about19.62m/s

Step by step solution

01

Given information and assumptions

mass of the rope 1.0kg

mass of the robot m=2.0kg

Tension of the ropeT=40N

Let us consider the Free Body Diagram (FBD) for the robot to move up and down in the rope.

Weight of the robot be w鈬赌=mg

Tension of the rope beT鈬赌

02

Explanation

To calculate acceleration of robot to move downward

From FBD, following Newton's third law, the Fnet鈬赌=0

role="math" localid="1649705336120" T鈬赌=-w鈬赌mropea=-mga=-mgmropea=-2脳9.811=-19.62m2/s

By symmetry, the acceleration of robot to move upward is19.62m2/s

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 lower block in FIGURE CP7.53 is pulled on by a rope with a tension force of 20 N. The coefficient of kinetic friction between the lower block and the surface is 0.30. The coefficient of kinetic friction between the lower block and the upper block is also 0.30.

What is the acceleration of the 2.0 kg block?

While driving to work last year, I was holding my coffee mug in my left hand while changing the CD with my right hand. Then the cell phone rang, so I placed the mug on the flat part of my dashboard. Then, believe it or not, a deer ran out of the woods and on to the road right in front of me. Fortunately, my reaction time was zero, and I was able to stop from a speed of 20m/s in a mere 50m, just barely avoiding the deer. Later tests revealed that the static and kinetic coefficients of friction of the coffee mug on the dash are 0.50 and 0.30, respectively; the coffee and mug had a mass of 0.50kg; and the mass of the deer was 120kg. Did my coffee mug slide?

A small car is pushing a large truck. They are speeding up. Is the force of the truck on the car larger than, smaller than, or equal to the force of the car on the truck?

Your forehead can withstand a force of about 6.0kNbefore fracturing, while your cheekbone can withstand only about 1.3kN. Suppose a 140gbaseball traveling at 30m/sstrikes your head and stops in 1.5ms

a. What is the magnitude of the force that stops the baseball?

b. What force does the baseball exert on your head? Explain.

c. Are you in danger of a fracture if the ball hits you in the forehead? On the cheek?

An 80kgspacewalking astronaut pushes off a 640kgsatellite, exerting a 100Nforce for the role="math" localid="1649256566272" 0.50sit takes him to straighten his arms. How far apart are the astronaut and the satellite after 1min.

See all solutions

Recommended explanations on Physics Textbooks

Conservation of Energy and Momentum

Read Explanation

Translational Dynamics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Turning Points in Physics

Read Explanation

Kinematics Physics

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.