Problem 84 - College Physics

Question

\(\bullet$$\bullet\) Some sliding rocks approach the base of a hill with a speed of 12 \(\mathrm{m} / \mathrm{s}\) . The hill rises at \(36^{\circ}\) above the horizontal and has coefficients of kinetic and static friction of 0.45 and 0.65 . respectively, with these rocks. Start each part of your solution to this problem with a free-body diagram. (a) Find the acceler- ation 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 there, show why. If it slides down, find its accel- cration oa the way down.

Step-by-Step Solution

Answer

(a) Acceleration up is \(-0.887 \text{ m/s}^2\). (b) Rocks slide down with \(7.698 \text{ m/s}^2\).

Step 1: Analyze Forces Along the Hill Incline

First, we need to consider the forces acting on the rocks as they slide up the hill. The forces include gravity, normal force, and kinetic friction. The gravitational force can be split into two components: one parallel to the incline and the other perpendicular. The component of gravity acting along the incline is \( mg \sin \theta \), and the normal force \( N \) acts perpendicular to the incline, balancing the perpendicular component of gravity, which is \( mg \cos \theta \). The kinetic friction can be calculated as \( f_k = \mu_k N = \mu_k mg \cos \theta \).

Step 2: Write the Equation for Net Force and Acceleration

Use Newton's second law to calculate the acceleration. The net force along the incline is the difference between the gravitational force component along the incline and the kinetic friction force. Thus the equation is: \( m a = mg \sin \theta - \mu_k mg \cos \theta \). Simplifying, we find the acceleration: \( a = g(\sin \theta - \mu_k \cos \theta) \). Substituting the known values, \( a = 9.8(\sin 36^\circ - 0.45 \cos 36^\circ) \). Calculate this to find \( a \).

Step 3: Calculate Numerical Value of Acceleration Up the Hill

Substitute the trigonometric values: \( \sin 36^\circ \approx 0.588 \) and \( \cos 36^\circ \approx 0.809 \). Calculate the acceleration: \( a = 9.8(0.588 - 0.45 \times 0.809) \). Solving this gives \( a \approx 0.887 \text{ m/s}^2 \) decelerating (note that the negative indicates deceleration as rocks are moving upward).

Step 4: Analyze Forces at the Highest Point

Once the rocks stop moving upward, analyze if they will start sliding down. At the highest point, there's no kinetic friction until they start moving again. To stay in place, the static friction must counterbalance the gravitational force component along the incline. The maximum static friction is \( f_s = \mu_s N = \mu_s mg \cos \theta \).

Step 5: Decide if the Rocks Slide or Stay in Place

The condition for the rocks to stay is if the gravitational component \( mg \sin \theta \) is less than or equal to the maximum static friction \( \mu_s mg \cos \theta \). Calculate to see: \( mg \sin 36^\circ < 0.65 \times mg \cos 36^\circ \). Simplifying gives \( \tan 36^\circ > 0.65 \), which is false since \( \tan 36^\circ \approx 0.727 < 0.65 \). Thus, rocks will slide down.

Step 6: Calculate Acceleration as Rocks Slide Down

As the rocks slide back down, calculate the acceleration again. This time, the friction force and gravitational force component both act in the same direction. The equation becomes: \( m a = mg \sin \theta - \mu_k mg \cos \theta \), using \( \mu_k \) for kinetic friction. Thus \( a = g(\sin \theta + \mu_k \cos \theta) \). Calculate: \( a = 9.8(0.588 + 0.45 \times 0.809) \), giving \( a \approx 7.698 \text{ m/s}^2 \).

Key Concepts

Kinetic Friction

When objects are in motion against each other, kinetic friction plays a vital role. In this problem, where rocks are sliding uphill on an inclined plane, kinetic friction opposes their movement. It is calculated using the formula:

• \( f_k = \mu_k N \)
• \( N \) is the normal force, exerted perpendicular to the contact surface.
• \( \mu_k \) represents the coefficient of kinetic friction, which is 0.45 in this scenario.

The normal force in this case is influenced by gravity. Specifically, it is given as \( mg \cos \theta \), where \( m \) stands for mass and \( \theta \) is the angle of inclination (36 degrees here). Hence, kinetic friction can be framed as:

• \( f_k = \mu_k mg \cos \theta \)

This acts to decelerate the rocks as they move uphill. Understanding kinetic friction is essential since it impacts the net force and, consequently, the rocks' acceleration.

Static Friction

Static friction is crucial when determining if the rocks will remain stationary at the hill's highest point. Static friction prevents movement until the external force exceeds it.

• It is calculated as \( f_s = \mu_s N \), where \( \mu_s \) is the coefficient of static friction.
• In this context, \( \mu_s \) is 0.65.

To decide if the rocks stay put at the peak, compare the static friction with the downhill gravitational force component:

• The gravitational force working to pull the rocks down is \( mg \sin \theta \).
• If \( mg \sin \theta \leq \mu_s mg \cos \theta \), the rocks will stay.

For our problem, calculations reveal \( \tan 36^\circ < 0.65 \), signalling that static friction is not enough to prevent the rocks from sliding. Knowing when static friction can maintain equilibrium is invaluable in solving these physics problems.

Newton's Second Law

Newton's Second Law of Motion is key to finding the acceleration of the rocks on the incline. It states that the acceleration \( a \) of an object is directly proportional to the net force \( F_{net} \) acting on it and inversely proportional to its mass \( m \):

• \( F_{net} = ma \)

For the rocks moving up the incline, the net force is determined by subtracting the kinetic friction (\( f_k \)) from the gravitational force component along the incline (\( mg \sin \theta \)). This derivation is expressed as:

• \( ma = mg \sin \theta - \mu_k mg \cos \theta \)
• By simplifying, acceleration can be expressed as \( a = g(\sin \theta - \mu_k \cos \theta) \)

This calculation assumes gravity is the dominating factor, compounded by the inclination. Newton’s law not only aids in determining acceleration but also offers a broader perspective on how objects move in response to varying forces.

Free-Body Diagram

Drawing free-body diagrams is an essential step in solving physics problems regarding forces. These diagrams illustrate all the forces acting on an object, like our rocks on an inclined plane.
To start, visualize and label each force:

• Gravity \( mg \), acting downwards through the center of mass.
• Normal force \( N \), perpendicular to the surface.
• Kinetic friction \( f_k \), opposing the motion of the rocks.

By breaking down the gravitational force into components, we understand better how each force influences the object:

• Component along the plane: \( mg \sin \theta \)
• Component perpendicular to the plane: \( mg \cos \theta \)

These forces collectively determine the movement of the rocks. Crafting a precise free-body diagram simplifies defining relationships between forces, eventually guiding how to apply formulas related to friction and Newton’s laws effectively.