Skip to main content
Ch 07: Newton's Third Law
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoKnight Calc 5th EditionCh 07: Newton's Third LawProblema 40b
Capitolo 7, Problema 40b

The 1.0 kg physics book in FIGURE P7.40 is connected by a string to a 500 g coffee cup. The book is given a push up the slope and released with a speed of 3.0 m/s. The coefficients of friction are μs = 0.50 and μk = 0.20. At the highest point, does the book stick to the slope, or does it slide back down?

Guida verificata passo dopo passo
1
Identify the forces acting on the book and the coffee cup. For the book, the forces include gravity (acting down the slope), the normal force (perpendicular to the slope), friction (opposing motion), and the tension in the string. For the coffee cup, the forces include gravity (acting downward) and the tension in the string (acting upward).
Determine the maximum static friction force that can act on the book. The formula for static friction is: fmax=μsN, where μs is the coefficient of static friction and N is the normal force. The normal force can be calculated as N=mgcosθ, where m is the mass of the book, g is the acceleration due to gravity, and θ is the angle of the slope.
Compare the forces acting on the book at the highest point. At this point, the book is momentarily at rest, so the static friction force must balance the component of gravity pulling the book down the slope and the tension in the string pulling it up. Write the force balance equation: fstatic=mbookgsinθ-T, where T is the tension in the string.
Calculate the tension in the string due to the coffee cup. The tension is equal to the weight of the coffee cup: T=mcupg, where mcup is the mass of the coffee cup.
Compare the static friction force required to keep the book stationary with the maximum static friction force. If the required static friction force exceeds the maximum static friction force, the book will slide back down the slope. Otherwise, it will stick to the slope.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.
Durata del video:
11m

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Friction

Friction is the force that opposes the relative motion of two surfaces in contact. It is characterized by two coefficients: static friction (μₛ), which prevents motion, and kinetic friction (μₖ), which acts when surfaces are sliding against each other. In this scenario, the static friction will determine whether the book can remain at rest at the highest point after being pushed up the slope.
Video consigliato:
Percorso guidato
08:11
Static Friction & Equilibrium

Energy Conservation

The principle of energy conservation states that energy cannot be created or destroyed, only transformed from one form to another. In this problem, the kinetic energy of the book when it is pushed up the slope will convert into gravitational potential energy as it rises, and any energy lost to friction will affect whether it can reach the highest point without sliding back down.
Video consigliato:
Percorso guidato
06:24
Conservation Of Mechanical Energy

Newton's Second Law

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass (F = ma). This law is crucial for analyzing the forces acting on the book as it moves up the slope, including gravitational force, frictional force, and the tension in the string connected to the coffee cup.
Video consigliato:
Percorso guidato
06:54
Intro to Forces & Newton's Second Law
Pratica correlata
Domanda del libro di testo

FIGURE P7.47 shows a 200 g hamster sitting on an 800 g wedge-shaped block. The block, in turn, rests on a spring scale. An extra-fine lubricating oil having μs = μk = 0 is sprayed on the top surface of the block, causing the hamster to slide down. Friction between the block and the scale is large enough that the block does not slip on the scale. What does the scale read, in grams, as the hamster slides down?

1128
views
Domanda del libro di testo

A 75 kg archer on ice skates is standing at rest on very smooth ice. He shoots a 450 g arrow horizontally. When released, the arrow reaches a speed of 110 m/s in 0.25 s. Assume that the force of the bow string on the arrow is constant. What is the archer's recoil speed?

1910
views
Domanda del libro di testo

In FIGURE CP7.54, find an expression for the acceleration of m1. The pulleys are massless and frictionless. Hint: Think carefully about the acceleration constraint.

1875
views
Domanda del libro di testo

A house painter uses the chair-and-pulley arrangement of FIGURE P7.45 to lift himself up the side of a house. The painter's mass is 70 kg and the chair's mass is 10 kg. With what force must he pull down on the rope in order to accelerate upward at 0.20 m/s².

3142
views
2
rank
Domanda del libro di testo

Your forehead can withstand a force of about 6.0 kN before fracturing, while your cheekbone can withstand only about 1.3 kN. Suppose a 140 g baseball traveling at 30 m/s strikes your head and stops in 1.5 ms. What is the magnitude of the force that stops the baseball?

2740
views
Domanda del libro di testo

A 2.0 kg block on a horizontal, frictionless surface is connected by a massless spring and a massless, frictionless pulley to a hanging mass. For what value of the hanging mass does the block accelerate at 1.5 m/s²?

609
views