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Some sliding rocks approach the base of a hill with a speed of 12 m/s. The hill rises at 36\(^\circ\) above the horizontal and has coefficients of kinetic friction and static friction of 0.45 and 0.65, respectively, with these rocks. (a) Find the acceleration of the rocks as they slide up the hill. (b) Once a rock reaches its highest point, will it stay there or slide down the hill? If it stays, show why. If it slides, find its acceleration on the way down.

Short Answer

Expert verified
(a) Acceleration up is -0.212 m/s². (b) Rock slides down with acceleration 3.46 m/s².

Step by step solution

01

Identify Forces on Rocks

First, identify the forces acting on the rocks as they slide up the hill. These include gravitational force, normal force, and frictional force. The gravitational force acting downward can be calculated as \( F_{gravity} = mg \sin(\theta) \), where \( m \) is the rock's mass, \( g \) is the acceleration due to gravity \( 9.8 \, \text{m/s}^2 \), and \( \theta \) is the incline angle, which is \( 36^\circ \). The normal force is perpendicular to the hill and can be calculated as \( F_{normal} = mg \cos(\theta) \). The frictional force, acting against motion, uses the coefficient of kinetic friction \( \mu_k = 0.45 \) and the normal force: \( F_{friction} = \mu_k \cdot F_{normal} \).
02

Calculating Frictional Force

Substitute to find the normal and frictional forces:\[ F_{normal} = mg \cos(36^\circ) \]\[ F_{friction} = 0.45 \times mg \cos(36^\circ) \]
03

Apply Newton's Second Law

Set the net force equation along the incline. The net force is the difference between the gravitational component along the incline and the frictional force:\[ F_{net} = mg \sin(36^\circ) - 0.45 \times mg \cos(36^\circ) \]Using Newton's Second Law, \( F_{net} = ma \). Solve for the acceleration \( a \):\[ a = g \sin(36^\circ) - 0.45g \cos(36^\circ) \]
04

Calculate Numerical Acceleration Up Hill

Calculate the numeric value of the acceleration: \[ a = 9.8 \cdot \sin(36^\circ) - 0.45 \cdot 9.8 \cdot \cos(36^\circ) \approx -0.212 \text{ m/s}^2 \]This negative sign indicates the rocks are decelerating.
05

Analyze Rock's Resting Condition

As the rock reaches its highest point, assess if it stays there. The maximum static friction force must be greater than or equal to the component of gravitational force pulling the rock back down for it to remain stationary:\[ F_{static,max} = \mu_s \cdot F_{normal} = 0.65 \cdot mg \cos(36^\circ) \]Compare this with gravitational component:\[ F_{gravity,\parallel} = mg \sin(36^\circ) \]. Since \( F_{static,max} < F_{gravity,\parallel}\), it cannot hold the rock. Therefore, the rock will slide down.
06

Determine Acceleration Down the Hill

The rocks will slide down when the static frictional force can no longer hold them. Two forces are acting when sliding down: gravitational component down the incline and kinetic friction opposing it. The net force:\[ F_{net,down} = mg \sin(36^\circ) - 0.45 \times mg \cos(36^\circ) \]Solve for acceleration \( a \):\[ a_{down} = g \sin(36^\circ) - 0.45g \cos(36^\circ) \approx 3.46 \text{ m/s}^2 \].

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Key Concepts

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

Newton's Second Law
Newton's Second Law is a fundamental principle of physics that relates the net force acting on an object to its mass and acceleration. The law is expressed by the equation \( F = ma \), where \( F \) is the net force acting on the object, \( m \) is the mass, and \( a \) is the acceleration. This equation shows that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass.
When applying Newton's Second Law to the exercise, the forces acting on the rocks must be identified. As they slide up the hill, these forces include gravitational force, the normal force, and frictional force. The goal is to find the net force in the direction parallel to the hill's incline, then use it to calculate the acceleration. The same principle applies for calculation when the rocks slide back down the hill.
gravitational force
Gravitational force is the force with which the Earth attracts a body towards its center. It plays a crucial role when analyzing motion on an inclined plane. The gravitational force is the reason why objects have weight.
  • The component of gravitational force that acts along the slope is calculated through \( F_{gravity,\parallel} = mg \sin(\theta) \).
  • The gravitational force perpendicular to the hill contributes to the normal force, calculated as \( F_{normal} = mg \cos(\theta) \).
Each component plays a role in determining motion and frictional forces on the incline. Gravitational component parallel to the incline helps in moving the rock down while perpendicular component balances the normal force.
static friction
Static friction is a force that acts between surfaces to prevent motion. It is what keeps an object stationary until enough force is applied to overcome it. Static friction is generally higher than kinetic friction, which is why it's harder to get an object moving than to keep it moving.
  • Static friction can be calculated as \( F_{static,max} = \mu_s \cdot F_{normal} \), where \( \mu_s \) is the static friction coefficient.
  • This frictional force must be overcome for the object to start sliding. In the exercise context, the rock does not stay at rest since the static friction force is not enough to counter the gravitational pull down the hill.
The key is to compare the potential static friction to the gravitational force component to determine whether sliding will occur or not.
acceleration calculation
Calculating acceleration involves applying Newton's Second Law by setting up the equation for net force and solving for acceleration \( a \). For an inclined plane problem like in the exercise:
  • The net force as the rock slides up is \( F_{net} = mg \sin(36^\circ) - f_{friction} \), where friction opposes motion and is a function of normal force and kinetic friction coefficient.
  • Substitute known values to solve for \( a = g \sin(36^\circ) - 0.45g \cos(36^\circ) \). This gives the acceleration value while sliding up.
  • When sliding back down, the forces involved are similar but the direction changes, resulting in a different net force and thus a different acceleration.
The math breaks down to translating physical forces into numerical representations, providing a clear picture of the problem dynamics.

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