/*! 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 55 A machine at a post office sends... [FREE SOLUTION] | 91Ó°ÊÓ

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A machine at a post office sends packages out a chute and down a ramp to be loaded into delivery vehicles. (a) Calculate the acceleration of a box heading down a \(10.0^{\circ}\) slope, assuming the coefficient of friction for a parcel on waxed wood is \(0.100 .\) (b) Find the angle of the slope down which this box could move at a constant velocity. You can neglect air resistance in both parts.

Short Answer

Expert verified
The acceleration of the box heading down a \(10.0^{\circ}\) slope is approximately 1.70 m/s^2. The angle of the slope down which this box could move at a constant velocity is approximately \(5.71^{\circ}\).

Step by step solution

01

Identify the known variables

In this exercise, we know: 1. Angle of the slope,ѳ = \(10.0^{\circ}\) 2. Coefficient of friction, μ = 0.100
02

Calculate the force of gravity acting on the parcel

First, we need to find the force of gravity (Fg) acting on the box along the inclined plane. The force of gravity along the plane can be given by: Fg = m * g * sin(ѳ) However, since we don't have a mass (m) for the parcel, we can't find an exact value for Fg in this step. Thus, we will leave it as an expression in terms of mass (m) for now. Fg = m * g * sin(\(10.0^{\circ}\))
03

Calculate the force of friction acting on the parcel

Next, we need to find the force of friction (Ff), which can be given by: Ff = μ * Fn Where Fn is the normal force acting on the parcel due to the incline. Since the box is on an inclined plane, we know that: Fn = m * g * cos(ѳ) Again, we don't have a mass, so we will leave Fn as an expression in terms of mass (m): Fn = m * g * cos(\(10.0^{\circ}\)) Now, substituting this expression for Fn into the equation for Ff: Ff = 0.100 * m * g * cos(\(10.0^{\circ}\))
04

Calculate the net force acting on the parcel

Since the parcel is going down the incline, we will subtract the force of friction from the force of gravity to get the net force acting on the parcel: Fnet = Fg - Ff Substitute the expressions for Fg and Ff: Fnet = (m * g * sin(\(10.0^{\circ}\))) - (0.100 * m * g * cos(\(10.0^{\circ}\)))
05

Calculate the acceleration of the parcel

Now that we have the net force acting on the parcel, we can use Newton's second law to find the acceleration: Fnet = m * a Substituting the expression for Fnet: m * a = (m * g * sin(\(10.0^{\circ}\))) - (0.100 * m * g * cos(\(10.0^{\circ}\))) Now, since we are looking for acceleration (a), we can divide both sides by mass (m) to get: a = g * (sin(\(10.0^{\circ}\)) - (0.100 * cos(\(10.0^{\circ}\)))) Substitute the value of g (9.81 m/s^2) and the angle into the equation and calculate the acceleration: a = 9.81 * (sin(\(10.0^{\circ}\)) - (0.100 * cos(\(10.0^{\circ}\)))) a ≈ 1.70 m/s^2 Thus, the acceleration of the box heading down a \(10.0^{\circ}\) slope is approximately 1.70 m/s^2.
06

Finding the angle for constant velocity

For the box to have a constant velocity, the net force acting on the parcel should be zero. This means that the force of gravity (Fg) should be equal to the force of friction (Ff): Fg = Ff Substitute the expressions for Fg and Ff: m * g * sin(ѳ) = 0.100 * m * g * cos(ѳ) We can cancel mass (m) and gravitational acceleration (g) from both sides: sin(ѳ) = 0.100 * cos(ѳ) Now divide by cos(ѳ) to get the tangent of the angle: tan(ѳ) = 0.100 Then, find the angle by taking the arctangent of 0.100: ѳ = arctan(0.100) ѳ ≈ \(5.71^{\circ}\) So, the angle of the slope down which this box could move at a constant velocity is approximately \(5.71^{\circ}\).

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Key Concepts

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

Force of Gravity
Understanding the force of gravity on an inclined plane is crucial for analyzing the motion of objects. Gravity is a force that pulls objects towards the center of the Earth, and its effects become more complex when objects are placed on a slope.

When an object is on an inclined plane, the force of gravity acting along the slope can be expressed as \( F_g = m \cdot g \cdot \sin(\theta) \), where \( m \) is the mass, \( g \) is the acceleration due to gravity (9.81 m/s2), and \( \theta \) is the angle of the incline with respect to the horizontal. This component of the gravitational force is what causes the object to accelerate down the slope, and it's a key factor in calculating the overall motion of the object.
Coefficient of Friction
The coefficient of friction (\( \mu \)) is a critical concept that describes the amount of frictional resistance between two surfaces. Frictional force can be found using the equation \( F_f = \mu \cdot F_n \), where \( F_f \) represents the force of friction and \( F_n \) is the normal force—the force perpendicular to the contact surface. On an incline, the normal force is less than the object's weight and is calculated as \( F_n = m \cdot g \cdot \cos(\theta) \).

The coefficient of friction does not have a unit as it is a ratio, providing a dimensionless value representing the ratio of the force of friction between two bodies and the force pressing them together. It's essential to consider both the coefficient of friction and the normal force to determine the force of friction acting on an object on an inclined plane.
Newton's Second Law
Newton's second law of motion is one of the cornerstones of classical mechanics. It states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass, formulated as \( F_{net} = m \cdot a \).

On an inclined plane, this law helps us calculate the object's acceleration by considering the net force, which includes gravitational force minus any frictional resistance. In the context of our example, once we determine the net force exerted on the package, we can calculate its acceleration by rearranging the law to solve for \( a \): \( a = \frac{F_{net}}{m} \). This profound relationship enables us to predict how the package will move down the chute, based on its mass and the forces acting upon it.
Constant Velocity
Constant velocity on an inclined plane is a special scenario where the forces acting on an object are balanced, resulting in zero acceleration. This means that the component of gravitational force down the slope is exactly counteracted by the force of friction.

For an object to move with a constant velocity, the net force must be zero: \( F_g = F_f \). This equilibrium means that the object will continue to move at the same speed—not speeding up or slowing down. By calculating the angle at which the forces balance each other, we can find the specific slope where an object, such as our parcel, will descend at a constant velocity. This is particularly useful in designing systems like chutes or ramps, where control over the speed of moving objects is desired.

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