Problem 30 - University Physics with Modern Physics

Question

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. (a) Find the acceleration 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 acceleration on the way down.

Step-by-Step Solution

Answer

(a) The acceleration up the hill is \(-6.14 \text{ m/s}^2\). (b) The rock will slide down with acceleration of \(2.66 \text{ m/s}^2\).

Step 1: Identify forces acting on the rocks

The forces acting on the rocks are gravity, friction, and the normal force. The gravitational force component parallel to the hill is given by \(mg \sin \theta\) and perpendicular to the hill by \(mg \cos \theta\), where \(m\) is the mass of the rock, \(g\) is the acceleration due to gravity (9.8 m/s²), and \(\theta = 36^{\circ}\). Kinetic friction is given by \(f_k = \mu_k \cdot N\), where \(\mu_k\) is the coefficient of kinetic friction and \(N = mg \cos \theta\).

Step 2: Calculate normal force

The normal force \(N\) is the component of gravitational force perpendicular to the hill. It can be computed as:\[ N = mg \cos \theta \]

Step 3: Determine frictional force upwards

The force of kinetic friction opposing motion as the rocks slide up is:\[ f_k = \mu_k \cdot N = \mu_k \cdot mg \cos \theta \]Substitute \(\mu_k = 0.45\), \(g = 9.8 \mathrm{m/s^2}\), and \(\theta = 36^{\circ}\).

Step 4: Calculate net force and acceleration up the hill

The net force acting on the rock along the hill's surface is:\[ F_{ ext{net, up}} = -f_k - mg \sin \theta \push into\rightarrow F_{ ext{net, up}} = ma \\]Thus, the acceleration \(a\) is:\[ a = \frac{-f_k - mg \sin \theta}{m} = -(\mu_k g \cos \theta + g \sin \theta) \]

Step 5: Determine if the rock stays at the top

Check static friction: the static frictional force \(f_s = \mu_s N\) must be greater than or equal to the gravitational force pulling the rock down (\(mg \sin \theta\)) for it to remain at its highest point. Substitute \(\mu_s = 0.65\):\[ f_s = \mu_s \cdot mg \cos \theta \geq mg \sin \theta \]

Step 6: Calculate if it slides back and find acceleration

If static friction is insufficient: repeat calculation considering the net force will cause it to slide back, where:\[ F_{ ext{net, down}} = mg \sin \theta - f_k \a_{ ext{down}} = \frac{mg \sin \theta - f_k}{m} = g \sin \theta - \mu_k g \cos \theta \]

Key Concepts

Static Friction

Static friction is a force that prevents objects from moving when they are at rest. It acts parallel to the surface of contact.
This force must be overcome for an object to begin moving. When dealing with slopes or inclined planes, like in this exercise, static friction plays a crucial role in determining whether a rock will stay put after coming to a stop.

In the scenario given, we use static friction to see if it can hold the rock in place once it reaches the highest point on the incline.
The formula to determine static friction is \( f_s = \mu_s \cdot N \), where \( \mu_s \) is the coefficient of static friction, and \( N \) is the normal force. Here \( N \) is calculated by \( mg \cos \theta \). This means the rock will remain stationary if static friction is greater than the downward gravitational force \( mg \sin \theta \). This balance is essential.

If the static frictional force is not sufficient to counteract the component of gravity pulling the rock back down the hill, the rock will start sliding back. It's the calculated friction that informs us if the incline and static force are enough to secure the rock or if it'll roll back on its way.

Acceleration

Acceleration is the rate of change of velocity of an object. On an inclined plane, several forces contribute to the acceleration of an object such as gravity and friction.
Calculating these forces helps to determine how quickly something moves up or down a slope.

In this exercise, we consider how acceleration affects rocks sliding up and potentially back down the hill. Using Newton’s second law, we find acceleration by the formula: \( a = \frac{F_{\text{net}}}{m} \). When rocks travel up, forces such as kinetic friction and gravitational pull slow them down, leading to negative acceleration. The acceleration equation becomes \(- (\mu_k g \cos \theta + g \sin \theta)\), indicating that resistance forces act opposite to the movement.

Should the rocks reach the top and not remain stationary due to inadequate static friction, they'll experience acceleration back downhill. The net force, minus the opposing frictional force, dictates this downward acceleration, found by \( g \sin \theta - \mu_k g \cos \theta \). Understanding acceleration in these terms helps in explaining motion on slopes.

Inclined Plane

An inclined plane is a flat surface tilted at an angle to the horizontal. It essentially transforms vertical motion into horizontal motion, making movement easier as opposed to lifting straight up.
This simple machine helps us study how forces like gravity and friction interact. When looking at an inclined plane, the angle of inclination greatly affects the forces at play and subsequently the motion of objects.

In this case, the hill forms an inclined plane, offering a welcoming example to observe how forces cause objects like rocks to move up or down. Any object on an incline experiences gravity pulling it down. This force is split into two components: one parallel (\( mg \sin \theta \)) and another perpendicular (\( mg \cos \theta \)) to the surface.
Here, friction, both static and kinetic, oppose this motion. It's crucial because these uphill and downhill forces determine whether an object accelerates or stays put.

Working with inclined planes involves understanding how to resolve forces into components and how those forces interact to create motion, or perhaps motion resistance, across the plane's surface.