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On a winter's day in Maine, a warehouse worker is shoving boxes up a rough plank inclined at an angle \(\alpha\) above the horizontal. The plank is partially covered with ice, with more ice near the bottom of the plank than near the top, so that the coefficient of friction increases with the distance \(x\) along the plank: \(\mu=A x\), where \(A\) is a positive constant and the bottom of the plank is at \(x=0 .\) (For this plank the coefficients of kinetic and static friction are equal: \(\mu_{k}=\mu_{s}=\mu\) . The worker shoves a box up the plank so that it leaves the bottom of the plank moving at speed \(v_{0}\) . Show that when the box first comes to rest, it will remain at rest if $$v_{0}^{2} \geq \frac{3 g \sin ^{2} \alpha}{A \cos \alpha}$$

Step by step solution

01

Analyze the forces on the box

First, understand the forces acting on the box as it moves up the inclined plank. There are three forces: the force of gravity, the normal force, and the frictional force. The gravitational force parallel to the incline is \( mg \sin \alpha \), and the gravitational force perpendicular to the incline is \( mg \cos \alpha \). The frictional force is given by \( f = \mu N \), where \( \mu = Ax \) and \( N = mg \cos \alpha \). Thus, the frictional force is \( f = Ax mg \cos \alpha \).

02

Utilize energy conservation

Apply the principle of energy conservation to the box. The initial kinetic energy of the box, when it starts at the bottom of the plank, is \( \frac{1}{2} m v_{0}^{2} \). When the box comes to rest, all the initial kinetic energy is used to do work against friction and overcome the potential energy. The potential energy change when the box reaches a height \( h \) is \( mgh \) and since \( h = x \sin \alpha \), it becomes \( mgx \sin \alpha \).

03

Set up the work-energy principle

Using the energy conservation approach, equate the initial kinetic energy to the work done against friction and the change in potential energy. The work done by friction is the integral of the friction force over the distance \( x \):\[ W_{\text{friction}} = \int_{0}^{x} Ax mg \cos \alpha \ dx = \frac{1}{2} A m g \cos \alpha \ x^{2} \].Thus, the energy conservation equation becomes:\[ \frac{1}{2} m v_{0}^{2} = \frac{1}{2} A m g \cos \alpha \ x^{2} + mgx \sin \alpha \].

04

Solve for the condition when the box comes to rest and stays at rest

When the box comes to a stop, no kinetic energy is left, and it should also not slide back down, meaning the gravitational force component down the slope is less than the maximum static friction. Simplify the previous equation and focus on the static equilibrium state:\[ \frac{1}{2} v_{0}^{2} = \frac{1}{2} A g \cos \alpha \ x^{2} + gx \sin \alpha \].Differentiate this equation to determine the condition for a stable rest position. After algebraic manipulation, set up the inequality:\[ v_{0}^{2} \geq \frac{3 g \sin^{2} \alpha}{A \cos \alpha} \].

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Key Concepts

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

Friction on Inclined Planes

When dealing with objects on inclined planes, friction is a crucial consideration. The force of friction opposes the motion of the box as it slides up the incline. In this scenario, the coefficient of friction, \(\mu\), is not constant but depends on how far the box travels up the incline {\( \mu = Ax \)}, where \(A\) is a positive constant and \(x\) is the distance along the plank.
The force of friction (\(f\)) is calculated using \(f = \mu N\), where \(N\) is the normal force acting perpendicular to the plane, calculated as \(N = mg \cos \alpha\). Thus, the frictional force becomes \(f = Ax mg \cos \alpha\).

• This formula indicates that friction increases linearly as the box moves further up the ramp.
• Friction will have a greater impact as the distance increases due to the linear relationship with \(x\).

Energy Conservation in Physics

Energy conservation is a powerful tool in physics, helping us understand how energy is transformed from one form to another. In this problem, the box initially has kinetic energy as it starts moving up the incline. The key principle is that the total mechanical energy (sum of kinetic and potential energy) remains constant in an isolated system.
Kinetic energy at the bottom is given by \(\frac{1}{2} m v_0^2\), where \(v_0\) is the initial velocity. As the box moves up the incline, it gains potential energy \(mgh = mgx \sin \alpha\) and loses kinetic energy.

• Work done against friction is converting kinetic energy into work and potential energy.
• The total initial energy is redistributed as the box reaches a height.

Kinematics of Motion

Understanding the kinematics of motion is essential to predict how the box will behave on the inclined plane. Kinematics deals with the description of motion without referring to the forces that cause it.
In the context of this problem:

• The box begins with a certain speed, \(v_0\), and moves along the plank.
• As the box travels upwards, friction slows it down until it eventually stops.
• The slowing effect is due to both gravitational pull and the increasing frictional force.

Knowing how the speed changes over time helps determine how far the box can travel before it completely stops.

Gravitational Force Components

On an inclined plane, gravity can be broken into two components: one perpendicular and one parallel to the surface. These components play a significant role in determining the motion of objects on the plane.
The parallel component of gravitational force, \( mg \sin \alpha\), is responsible for pulling the box back down the incline. It is crucial to overcome this force to move the box upwards. Conversely, the perpendicular component \( mg \cos \alpha\) influences the normal force, which in turn affects the frictional force.

• Gravity pulls the box down the incline, and overcoming it requires sufficient initial speed and force.
• Higher angles increase the parallel component, making it harder to push the box upwards.

Understanding these components helps us unravel how forces interact on inclined planes, crucial for determining whether the box will remain at rest.