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A package is projected up a \(15^{\circ}\) incline at \(A\) with an initial velocity of \(8 \mathrm{m} / \mathrm{s}\). Knowing that the coefficient of kinetic friction between the package and the incline is \(0.12,\) determine \((a)\) the maximum distance \(d\) that the package will move up the incline, \((b)\) the velocity of the package as it returns to its original position.

Short Answer

Expert verified
(a) The maximum distance is calculated using kinematics with energy loss. (b) The velocity upon return is influenced by friction and potential energy gain.

Step by step solution

01

Analyze the forces acting on the package

The forces acting on the package are gravity, normal force and friction. The gravitational force can be split into two components: one parallel to the incline \(mg \sin(\theta)\) and one perpendicular to it \(mg \cos(\theta)\). The friction force will act down the incline with magnitude \(f_k = \mu_k N = \mu_k mg \cos(\theta)\).
02

Set up the equations for motion up the incline

Apply Newton's second law in the direction parallel to the incline. The net force \(F_{net}\) is given by the initial kinetic energy minus work done by friction and gravity. Hence, \(m \cdot a = -mg \sin(\theta) - \mu_k mg \cos(\theta)\). Solve for acceleration \(a\):\[ a = -g (\sin(\theta) + \mu_k \cos(\theta)) \].
03

Determine the maximum distance up the incline using kinematics

Using the kinematic equation \(v^2 = u^2 + 2ad\) where final velocity \(v = 0\) at maximum distance, initial velocity \(u = 8\ \text{m/s}\), and \(a\) from Step 2. Solve for \(d\): \[ 0 = 8^2 + 2(-g(\sin(15^\circ) + 0.12\cos(15^\circ))d \]. Thus, \[ d = \frac{8^2}{2g(\sin(15^\circ) + 0.12\cos(15^\circ))} \].
04

Apply the work-energy principle for return motion

As the package slides back, it initially has zero velocity at the peak and returns to its initial position slightly accelerated by friction as well as gravity assist. Use work-energy again: \[ 0 + W_{net} = \frac{1}{2}mv^2 - 0 \] with friction and gravity acting to increase speed. Substitute \(W_{net} = mgh(2 \times Friction)\).
05

Calculate the return velocity

Now find the velocity \(v\) when the package returns to point \(A\) using energy equations: \[ v^2 = 0 + 2a\cdot d \]. Substitute \(a\) and \(d\) already determined to find final \(v\).

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Key Concepts

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

Kinetic Friction
Kinetic friction is a force that opposes the relative motion between two surfaces in contact as they slide past each other. In the context of an inclined plane, the kinetic friction plays a significant role. It acts opposite to the direction of motion. This frictional force is defined by the equation \( f_k = \mu_k N \), where \( \mu_k \) is the coefficient of kinetic friction and \( N \) is the normal force.

For an object on an incline, the normal force isn't equal to the object's weight. Instead, it's the perpendicular component of the object's weight relative to the incline surface. We can calculate it as \( mg \cos(\theta) \). Thus, the kinetic friction force acting on the incline would be \( \mu_k mg \cos(\theta) \). You should always remember that kinetic friction depends on both the nature of the surfaces in contact and how strongly the surfaces are pressed together.
Kinematics
Kinematics deals with the motion of objects without considering the causes of the motion. On an inclined plane, it helps to find parameters such as velocity, acceleration, and distance traveled. A common kinematic equation used is \( v^2 = u^2 + 2ad \), where \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( d \) is the distance traveled.

In our exercise, to find how far the package slides up the incline, we use this equation by setting \( v \) to zero because the package stops momentarily at its maximum height. By knowing the initial velocity and calculating the acceleration from other forces, we can determine the distance \( d \). It's crucial to understand how motion equations work together with forces, especially considering both the initial and final state of the motion.
Newton's Second Law
Newton's Second Law of Motion is fundamental for analyzing dynamics on an inclined plane. It states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration: \( F_{net} = ma \). This principle is instrumental in determining how the net forces act on the package on the incline.

In the problem, the net force is derived from various components: gravitational force along the incline, normal force, and kinetic friction force. The gravitational force parallel to the incline is given by \( mg \sin(\theta) \). By applying Newton's second law, we sum these forces to find the net force and then solve for acceleration. The direct relationship between forces and resulting acceleration helps us to analyze and predict the movement behavior of objects on slopes.
Work-Energy Principle
The work-energy principle connects the concepts of work, energy, and motion. It states that the work done on an object results in a change in its kinetic energy. When analyzing the package's return trip down the incline, energy transitions are crucial.

As the package goes back down, potential energy at the highest point converts back into kinetic energy, overcoming friction. The work done by gravity on the package, minus the work done by friction, will equal the change in kinetic energy. This principle can simplify calculations by emphasizing energy changes, rather than individual forces acting throughout the object’s path. Using this principle also confirms results from kinematic and dynamic equations, offering a comprehensive understanding of the motion.

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Most popular questions from this chapter

A 1.5 -lb ball that can slide on a horizontal frictionless surface is attached to a fixed point \(O\) by means of an elastic cord of constant \(k=1\) lb/n. and undeformed length 2 ft. The ball is placed at point \(A,\) 3 ft from \(O,\) and given an initial velocity \(v_{0}\) perpendicular to \(O A,\) allowing the ball to come within a distance \(d=9\) in. of point \(O\) after the cord has become slack. Determine \((a)\) the initial speed \(v_{0}\) of the ball, \((b)\) its maximum speed.

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