Skip to main content
Ch 06: Dynamics I: Motion Along a Line
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 06: Dynamics I: Motion Along a LineProblema 74b
Capitolo 6, Problema 74b

A block of mass m is at rest at the origin at t = 0. It is pushed with constant force F₀ from 𝓍 = 0 to 𝓍 = L across a horizontal surface whose coefficient of kinetic friction is μₖ = μ₀ ( 1 - 𝓍/L ) . That is, the coefficient of friction decreases from μ₀ at 𝓍 = 0 to zero at 𝓍 = L. b. Find an expression for the block's speed as it reaches position L.

Guida verificata passo dopo passo
1
Step 1: Begin by identifying the forces acting on the block. The block experiences a constant applied force F₀, a frictional force that depends on position 𝓍, and the normal force balancing the gravitational force. The frictional force is given by F_friction = μₖ * m * g, where μₖ = μ₀ * (1 - 𝓍/L).
Step 2: Write the net force acting on the block as a function of position 𝓍. The net force is F_net = F₀ - F_friction. Substituting the expression for F_friction, we get F_net = F₀ - μ₀ * (1 - 𝓍/L) * m * g.
Step 3: Use Newton's second law to relate the net force to the block's acceleration. Newton's second law states F_net = m * a, where a is the acceleration. Substituting F_net, we have m * a = F₀ - μ₀ * (1 - 𝓍/L) * m * g. Simplify to find the acceleration: a = (F₀/m) - μ₀ * g * (1 - 𝓍/L).
Step 4: To find the block's speed at position 𝓍 = L, use the work-energy principle. The work done by the net force is equal to the change in kinetic energy. The work done by the applied force and friction can be calculated by integrating the net force over the distance from 𝓍 = 0 to 𝓍 = L. The work-energy principle states: W_net = ΔK = (1/2) * m * v² - 0, where v is the speed at 𝓍 = L.
Step 5: Perform the integration to find the work done by the net force. The net force is F_net = F₀ - μ₀ * (1 - 𝓍/L) * m * g. Integrate F_net with respect to 𝓍 from 0 to L: W_net = ∫[0 to L] (F₀ - μ₀ * (1 - 𝓍/L) * m * g) d𝓍. Solve this integral to find W_net, then equate it to (1/2) * m * v² to solve for v, the block's speed at position L.

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:
14m

Concetti chiave

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

Newton's Second Law of Motion

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. This relationship is expressed mathematically as F = ma, where F is the net force, m is the mass, and a is the acceleration. Understanding this law is crucial for analyzing the motion of the block under the influence of the applied force and friction.
Video consigliato:
Percorso guidato
06:54
Intro to Forces & Newton's Second Law

Friction and its Coefficient

Friction is the force that opposes the relative motion of two surfaces in contact. The coefficient of kinetic friction (μₖ) quantifies this force and varies depending on the surfaces involved. In this scenario, the coefficient decreases linearly from μ₀ to zero as the block moves from position 0 to L, affecting the net force and, consequently, the block's acceleration and speed.
Video consigliato:
Percorso guidato
08:11
Static Friction & Equilibrium

Work-Energy Principle

The Work-Energy Principle states that the work done on an object is equal to the change in its kinetic energy. In this problem, the work done by the applied force and the work done against friction will determine the block's final speed at position L. By calculating the net work done on the block, one can derive an expression for its speed as it reaches the end of the distance L.
Video consigliato:
Percorso guidato
04:10
The Work-Energy Theorem
Pratica correlata
Domanda del libro di testo

A 4.0-cm-diameter, 55 g ball is shot horizontally into a tank of 40°C honey. How long will it take for the horizontal speed to decrease to 10% of its initial value?

1465
views
Domanda del libro di testo

A 1.0-cm-diameter, 2.0 g marble is shot horizontally into a tank of 20°C olive oil at 10 cm/s. How far in cm will it travel before stopping?

1321
views
Domanda del libro di testo

A 500 g ball moves horizontally with velocity v𝓍 = ( 15 m) / (t + 1 s) for t > 0 s. What is the net force on the ball at t = 1 s?

352
views
Domanda del libro di testo

A ball is shot from a compressed-air gun at twice its terminal speed. What is the ball's initial acceleration, as a multiple of g, if it is shot straight up?

2113
views
Domanda del libro di testo

What is the magnitude of the acceleration of a skydiver at the instant she is falling at one-half her terminal speed?

1898
views
Domanda del libro di testo

A spherical particle of mass m is shot horizontally with initial speed v₀ into a viscous fluid. Use Stokes' law to find an expression for vₓ (t), the horizontal velocity as a function of time. Vertical motion due to gravity can be ignored.

1497
views