Problem 102 - University Physics with Modern Physics

Question

CALC On a winter 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: \(\mu_{\mathrm{k}}=\mu_{\mathrm{s}}=\mu .\) The worker shoves a box up the plank so that it leaves the bottom of the plank mov- ing 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

Answer

The box remains at rest if \( v_0^2 \geq \frac{3g \sin^2 \alpha}{A \cos \alpha} \).

Step 1: Understanding the Problem

The problem involves a box sliding up a plank with a variable coefficient of friction. We need to show that when the box stops, it remains at rest under certain conditions. The key is to relate the given kinetic energy condition as a balance of forces and energy on the incline, using the friction coefficient function \( \mu = Ax \).

Step 2: Determining the Forces on the Box

The forces are: gravitational force \( mg \) acting downwards, normal force \( N \) acting perpendicular to the plank, and frictional force \( f = \mu N = AxN \) acting opposite to the box's motion.Thus, the normal force \( N = mg \cos \alpha \).

Step 3: Calculating Work Done Against Friction

The work done by friction as the box moves to position \( x \) is given by \[ W_f = \int_0^x (Ax)(mg \cos \alpha) \, dx = mg \cos \alpha \cdot \frac{Ax^2}{2} \].

Step 4: Energy Conservation Equation

Initially, the box has kinetic energy \( \frac{1}{2}mv_0^2 \). When it comes to rest, this energy equals the work done against friction:\[ \frac{1}{2}mv_0^2 = mg \cos \alpha \cdot \frac{Ax_c^2}{2} \ \implies v_0^2 = g \cos \alpha \cdot Ax_c^2 \].

Step 5: Solving for Critical Condition

Use the condition when box comes to rest and cannot slide back:\( \mu_{s} mg \cos \alpha > mg \sin \alpha \). Given \( \mu = Ax \), substitute \( x_c \) to find:\[ Ax_c mg \cos \alpha > mg \sin \alpha \ \Rightarrow x_c > \frac{\sin \alpha}{A \cos \alpha} \].

Step 6: Comparing Stopping Distance and Energy Condition

From the energy equation \( x_c^2 = \frac{v_0^2}{g \cos \alpha A} \). Combining with the above, substitute back:\[ \frac{v_0^2}{g \cos \alpha A} \geq \left(\frac{\sin \alpha}{A \cos \alpha}\right)^2 \ \Rightarrow v_0^2 \geq \frac{3g \sin^2 \alpha}{A \cos \alpha} \].

Step 7: Conclusion

The derivation aligns with the energy and static friction condition required for the box to remain at rest after stopping on the incline. The box stays at rest if the initial velocity condition is met, which confirms the problem's statement.

Key Concepts

Variable Coefficient of Friction

A fascinating aspect of this problem is the variable coefficient of friction, denoted as \( \mu = Ax \). Interestingly, this approach means that the friction experienced by the box changes as it moves up the plank.
Initially, at the bottom of the plank where \( x = 0 \), the coefficient of friction is practically nonexistent. As the box travels further, friction increases linearly with distance due to \( Ax \).
This variability simulates real-life scenarios, such as icy roads that get progressively less slippery. The concept highlights how friction isn't always constant, and understanding this can help predict movement in variable conditions.

• This feature is crucial for calculating the correct stopping point, as it requires integrating friction over distance.
• In this problem, it affects how much work friction can do as the box slides up the inclined plane.

Understanding the variable coefficient of friction is key to predicting how the box behaves on different parts of the plank.

Energy Conservation

Energy conservation is the critical principle that helps us solve this problem. Initially, the box has a kinetic energy determined by its mass \( m \) and velocity \( v_0 \). The kinetic energy can be calculated using the formula \( \frac{1}{2}mv_0^2 \).
As the box moves up the incline, friction does work against it. This work done decreases the box's kinetic energy until it comes to rest. Thus, the initial kinetic energy transforms into work done against friction. The relationship is expressed by the equation: \[ \frac{1}{2}mv_0^2 = mg \cos \alpha \cdot \frac{Ax_c^2}{2} \]
This balancing act ensures that when the box has exhausted all its kinetic energy through work done against friction, it stops moving. It's a compelling demonstration of how energy changes form but overall remains conserved in a closed system.

Static and Kinetic Friction Equivalence

The problem simplifies by assuming that static friction \( \mu_s \) and kinetic friction \( \mu_k \) are the same, both represented as \( \mu = Ax \). This equivalence is not typical, as generally, static friction is greater than kinetic friction.
In scenarios where they are equal, the transition between when the object starts moving and when it stops doesn’t involve a sudden change in frictional force. It simplifies calculations as we do not need to adjust frictional values mid-problem.
For this exercise, it means friction is always calculated using \( Ax \). The advantage is a seamless switch from movement to rest, making calculations straightforward.

• This equivalence allows us to apply the same friction coefficient to calculate stopping conditions.
• It ensures that once the box stops, the forces acting on it are sufficient to keep it at rest, provided our derived condition is met.

Overall, this assumption helps highlight the roles of \( \alpha \) and \( v_0 \) without dwelling too much on the friction phase transition.

Inclined Plane Dynamics

Understanding the dynamics of an inclined plane is crucial for solving problems like this. Here, the incline angle \( \alpha \) greatly influences the forces acting on the box.
The box experiences a gravitational component \( mg \sin \alpha \), pulling it down the incline, and a normal force \( N = mg \cos \alpha \), acting perpendicular to the surface.
Additionally, frictional force \( f = \mu N \) acts opposite the box’s motion. The incline angle affects both friction and motion, adjusting how these forces’ interplay.
Forces on inclined planes are managed via:

• Normal force
  This acts perpendicular to the surface, affecting the maximum possible friction.
• Gravitational pull
  This component strives to slide the box downward.

By combining these forces properly, one predicts the box's motion and checks if it will stay put once stopping, based on energy and friction calculations.
This planning enables detailed insight into how an inclined surface modifies an everyday motion scenario, underpinning many practical applications in physics.