Problem 48 - University Physics with Modern Physics

Question

Up and Down the Hill. A \(28-\mathrm{kg}\) rock approaches the foot of a hill with a speed of 15 \(\mathrm{m} / \mathrm{s}\) . This hill slopes upward at a constant angle of \(40.0^{\circ}\) above the horizontal. The coefficients of static and kinetic friction between the hill and the rock are 0.75 and 0.20 , respectively. (a) Use energy conservation to find the maximum height above the foot of the hill reached by the rock. (b) Will the rock remain at rest at its highest point, or will it slide back down the hill? (c) If the rock does slide back down, find its speed when it returns to the bottom of the hill.

Step-by-Step Solution

Answer

Maximum height is 5.78 m, rock stays at rest due to static friction.

Step 1: Understand the Problem

We have a 28 kg rock moving up a hill with an initial speed of 15 m/s. The hill is inclined at an angle of \(40^\circ\). We need to use energy conservation to find the maximum height reached, check if the rock remains at rest, and find its speed when it returns.

Step 2: Calculate Initial Kinetic Energy

The initial kinetic energy (KE) of the rock can be calculated using the formula \( KE = \frac{1}{2} m v^2 \). Here, \( m = 28 \, \text{kg} \) and \( v = 15 \, \text{m/s} \). Thus, \( KE = \frac{1}{2} \times 28 \times (15)^2 = 3150 \, \text{J} \).

Step 3: Determine Energy Lost to Friction

The friction force acting against the rock's motion as it ascends is \( f_k = \mu_k mg \cos(\theta) \), where \( \mu_k = 0.20 \) is the coefficient of kinetic friction. The work done against friction \( W_f = f_k \times d \), where \( d \) is the distance traveled. We will calculate \( d \) when finding the height.

Step 4: Calculate Work Against Friction with Energy Perspective

Since \( h = d \sin(\theta) \), the energy conservation equation, accounting for work done against friction, is \( KE_{\text{initial}} - W_f = mgh \). Rearranging and finding \( d = \frac{2 KE}{m g \sin(\theta) + \mu_k mg \cos(\theta)} \), derive final formulas for \( h = d \sin(\theta) \).

Step 5: Maximum Height Calculation

Using algebra and substituting known values \( KE = 3150 \, \text{J} \), \( \sin(40^\circ) = 0.6428 \), and \( \cos(40^\circ) = 0.7660 \): \ \( h = \frac{2 \times 3150}{28 \times 9.8 \times 0.6428 + 0.20 \times 28 \times 9.8 \times 0.7660} = 5.78 \, \text{m} \).

Step 6: Determine if Rock Slides Back

To see if the rock stays at rest, check static friction. If \( \mu_s mg \cos(\theta) \geq mg \sin(\theta) \), the rock doesn't slide. Calculate \( 0.75 \times mg \cos(\theta) = 207.59 \, \text{N} \) and \( mg \sin(\theta) = 178.42 \, \text{N} \). Since static friction exceeds gravitational pull, the rock stays at rest.

Step 7: Speed Calculation if Rock Slides

Since the rock stays at rest according to static friction analysis, parts b and c confirm no need to calculate return speed as it doesn't slide back down.

Key Concepts

Kinetic Friction

When a body, like our rock, moves up an inclined plane, it experiences kinetic friction. This type of friction occurs between surfaces in relative motion. The force of friction acts opposite to the direction of movement, slowing the rock down and converting some of its mechanical energy into thermal energy.

The force due to kinetic friction can be calculated using the formula:

• \( f_k = \mu_k \cdot N \)

where \( \mu_k \) is the coefficient of kinetic friction, and \( N = mg \cos(\theta) \) is the normal force. The normal force is the component of the rock's weight perpendicular to the inclined plane.

Understanding kinetic friction is crucial because it determines how much energy is lost as the rock moves up. In our problem, this friction uses some of the rock's initial kinetic energy, reducing the height the rock can reach before stopping. Recognizing this helps us apply energy conservation properly to find the maximum height.

Static Friction

Static friction is essential to explore when considering if the rock will slide back down the hill once it stops. Unlike kinetic friction, static friction prevents motion. Its magnitude varies up to a maximum value given by:

• \( f_s \leq \mu_s \cdot N \)

where \( \mu_s \) is the coefficient of static friction, and \( N = mg \cos(\theta) \) is the normal force. The condition \( \mu_s \cdot N \geq mg \sin(\theta) \) implies that the maximum static frictional force is enough to counter the downhill gravitational pull, stabilizing the rock at its peak position.

In our exercise, the static friction was sufficient to hold the rock at rest, as \( 207.59 \, \text{N} > 178.42 \, \text{N} \), so the rock didn't slide back. This comparison is pivotal because it compares the rock's potential energy against its tendency to roll back due to gravity.

Inclined Planes

Understanding inclined planes is central to problems involving motion on slopes. An inclined plane is a flat surface tilted at an angle to the horizontal. The angle causes gravitational force to split into two components:

• Parallel to the plane: \( mg \sin(\theta) \) - causes the object to slide down
• Perpendicular to the plane: \( mg \cos(\theta) \) - affects the normal force

These components are critical in both calculating kinetic and static friction, as well as assessing the work done by gravity and resistance forces.

By analyzing how these forces interact, we gain insights into how the initial kinetic energy translates into potential energy as the rock climbs the hill. The conservation of energy principle ties these forces and energy changes together, offering a comprehensive understanding of the rock's journey up and down the inclined plane.