/*! 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 105 As a \(40 \mathrm{~N}\) block sl... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

As a \(40 \mathrm{~N}\) block slides down a plane that is inclined at \(25^{\circ}\) to the horizontal, its acceleration is \(0.80 \mathrm{~m} / \mathrm{s}^{2}\), directed up the plane. What is the coefficient of kinetic friction between the block and the plane?

Short Answer

Expert verified
The coefficient of kinetic friction is approximately 0.376.

Step by step solution

01

Identify the Forces

First, we need to identify the forces acting on the block. The forces are: the gravitational force \(F_g\) acting vertically downward, the normal force \(F_n\) acting perpendicular to the inclined plane, and the frictional force \(F_k\) acting against the direction of motion up the plane. The gravitational force component along the plane is \(F_{gx} = mg \sin \theta\) and the normal force component is \(F_{gy} = mg \cos \theta\), where \(\theta = 25^\circ\), \( g = 9.8 \mathrm{~m/s^2} \), and \( m = \frac{F}{g} = \frac{40}{9.8} \mathrm{~kg}\).
02

Set Up the Equation of Motion

The net force acting along the plane direction (up the incline is positive) can be set up using Newton's second law: \(ma = F_{gx} - F_k\), and rearranging gives \(F_k = F_{gx} - ma\). Substituting the gravitational component, this becomes: \(F_k = mg \sin \theta - ma\).
03

Calculate Components of Forces

Calculate the gravitational component along the incline: \(F_{gx} = 40 \sin 25^\circ\approx 40 \times 0.4226 = 16.9\mathrm{~N}\). The acceleration component is: \(ma = \frac{40}{9.8} \times 0.80 = 3.27 \mathrm{~N}\).
04

Solve for Frictional Force

The frictional force is obtained by \(F_k = 16.9 - 3.27 = 13.63 \mathrm{~N}\).
05

Calculate the Normal Force

The normal force \(F_n\) can be found using the perpendicular component of gravity: \(F_n = mg \cos \theta = 40 \cos 25^\circ \approx 40 \times 0.9063 = 36.25 \mathrm{~N}\).
06

Determine the Coefficient of Kinetic Friction

The coefficient of kinetic friction \(\mu_k\) is defined by the ratio of the frictional force to the normal force: \(\mu_k = \frac{F_k}{F_n} = \frac{13.63}{36.25} \approx 0.376\).

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 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Newton's Second Law
Newton's second law of motion is a fundamental principle that describes how the motion of an object changes when forces act upon it. It can be stated as: "The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass." The equation that expresses this law is:
  • \( F = ma \)
where \( F \) is the net force in Newtons, \( m \) is the mass in kilograms, and \( a \) is the acceleration in meters per second squared.

In our exercise, the block has various forces acting on it as it slides down an inclined plane. To solve for its motion, Newton's second law is applied along the plane. The net force along the incline can be specifically written as the difference between the gravitational force component down the incline and the frictional force opposing the motion and the block's acceleration.
  • This results in the equation: \( ma = F_{gx} - F_k \)
By rearranging, we can find the frictional force \( F_k \). Newton's second law is a powerful tool for analyzing motion in physics, especially when multiple forces are involved.
Forces on an Incline
When an object is on an inclined plane, the forces operating on it need careful analysis to understand the motion correctly. The primary forces to consider are:
  • The gravitational force \( F_g \), which acts downwards.
  • The normal force \( F_n \), which acts perpendicular to the surface of the incline.
  • The frictional force \( F_k \), which opposes the direction of motion.

The gravitational force can be split into two components using trigonometric identities:
  • The component along the incline (parallel), \( F_{gx} = mg \sin \theta \).
  • The component perpendicular to the incline, \( F_{gy} = mg \cos \theta \).
In our problem, the inclined angle is \(25^\circ\) and these components are crucial for determining how the block moves. The component \( F_{gx} \) is what causes the block to slide down. The normal force \( F_n \), calculated from \( F_{gy} \), helps in determining the frictional force. Understanding these forces helps portray a clearer picture of the dynamics on an incline.
Coefficient of Friction
The coefficient of kinetic friction \( \mu_k \) is a dimensionless value that represents the frictional resistance between surfaces in motion relative to each other. It offers insight into how easily one surface can slide over another.

It's calculated using the formula:
  • \( \mu_k = \frac{F_k}{F_n} \)
where \( F_k \) is the kinetic frictional force and \( F_n \) is the normal force.

In our scenario, once the frictional force \( F_k \) is determined using the net force along the incline, and the normal force \( F_n \) is calculated based on the component of gravity, \( \mu_k \) can be directly computed. Recognizing the value of \( \mu_k \) helps physicists and engineers evaluate material properties and understand how surfaces interact under motion, which is vital for designing anything from machinery to everyday objects.

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

A child places a picnic basket on the outer rim of a merrygo-round that has a radius of \(4.6 \mathrm{~m}\) and revolves once every \(30 \mathrm{~s}\). (a) What is the speed of a point on that rim? (b) What is the lowest value of the coefficient of static friction between basket and merry-go-round that allows the basket to stay on the ride?

A \(1.5 \mathrm{~kg}\) box is initially at rest on a horizontal surface when at \(t=0\) a horizontal force \(\vec{F}=(1.8 t) \hat{\mathrm{i}} \mathrm{N}\) (with \(t\) in seconds) is applied to the box. The acceleration of the box as a function of time \(t\) is given by \(\vec{a}=0\) for \(0 \leq t \leq 2.8 \mathrm{~s}\) and \(\vec{a}=(1.2 t-2.4) \hat{\mathrm{i}} \mathrm{m} / \mathrm{s}^{2}\) for \(t>\) \(2.8 \mathrm{~s}\). (a) What is the coefficient of static friction between the box and the surface? (b) What is the coefficient of kinetic friction between the box and the surface?

A police officer in hot pursuit drives her car through a circular turn of radius \(300 \mathrm{~m}\) with a constant speed of \(80.0 \mathrm{~km} / \mathrm{h}\). Her mass is \(55.0 \mathrm{~kg}\). What are (a) the magnitude and (b) the angle (relative to vertical) of the net force of the officer on the car seat? (Hint: Consider both horizontal and vertical forces.)

\( \Rightarrow \sqrt{2}\) A sling-thrower puts a stone \((0.250 \mathrm{~kg})\) in the sling's pouch \((0.010 \mathrm{~kg})\) and then begins to make the stone and pouch move in a vertical circle of radius \(0.650 \mathrm{~m}\). The cord between the pouch and the person's hand has negligible mass and will break when the tension in the cord is \(33.0 \mathrm{~N}\) or more. Suppose the slingthrower could gradually increase the speed of the stone. (a) Will the breaking occur at the lowest point of the circle or at the highest point? (b) At what speed of the stone will that breaking occur?

\(\infty \Rightarrow \sqrt{\text { A roller-coaster car at an amusement park has a mass }}\) of \(1200 \mathrm{~kg}\) when fully loaded with passengers. As the car passes over the top of a circular hill of radius \(18 \mathrm{~m}\), assume that its speed is not changing. At the top of the hill, what are the (a) magnitude \(F_{N}\) and (b) direction (up or down) of the normal force on the car from the track if the car's speed is \(v=11 \mathrm{~m} / \mathrm{s} ?\) What are (c) \(F_{N}\) and (d) the direction if \(v=14 \mathrm{~m} / \mathrm{s}\) ?

See all solutions

Recommended explanations on Physics Textbooks

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.