Problem 38 - College Physics

Question

A toboggan approaches a snowy hill moving at \(11.0 \mathrm{~m} / \mathrm{s}\). The coefficients of static and kinetic friction between the snow and the toboggan are 0.40 and \(0.30,\) respectively, and the hill slopes upward at \(40.0^{\circ}\) above the horizontal. Find the acceleration of the toboggan (a) as it is going up the hill and (b) after it has reached its highest point and is sliding down the hill.

Step-by-Step Solution

Answer

Up the hill: 8.3 m/s², Down the hill: 3.5 m/s².

Step 1: Understanding Forces on the Toboggan

When the toboggan is moving, the forces acting on it include gravitational force, normal force, and frictional force. We need to find the frictional force using the coefficient of kinetic friction since the toboggan is moving.

Step 2: Calculating Gravitational Force Components

The gravitational force component along the slope is given by \( F_{g, \parallel} = mg \sin \theta \), and perpendicular to the slope, it is \( F_{g, \perp} = mg \cos \theta \), where \( m \) is the mass of the toboggan, and \( \theta \) is the angle of the slope.

Step 3: Calculating Frictional Force

The kinetic frictional force can be calculated using \( F_{f} = \mu_k F_{g, \perp} = \mu_k mg \cos \theta \), where \( \mu_k \) is the coefficient of kinetic friction \(0.30\).

Step 4: Determining Acceleration Up the Hill

The net force, when going up the hill, is \( F_{net} = F_{f} + F_{g, \parallel} \). Using \( F_{net} = ma \), solve for acceleration: \( a = \frac{F_{f} + F_{g, \parallel}}{m} = g(\mu_k \cos \theta + \sin \theta) \).

Step 5: Calculating Acceleration Down the Hill

For the toboggan sliding down, the net force changes direction: \( F_{net} = F_{g, \parallel} - F_{f} \). Solve for acceleration: \( a = \frac{F_{g, \parallel} - F_{f}}{m} = g(\sin \theta - \mu_k \cos \theta) \).

Step 6: Plug in Known Values and Solve

Using \( g = 9.8 \text{ m/s}^2\) and \( \theta = 40^ ext{°} \), calculate the accelerations: Up the hill \( a_{up} = 9.8(0.30 \, \cos 40^ ext{°} + \sin 40^ ext{°}) \), Down the hill \( a_{down} = 9.8(\sin 40^ ext{°} - 0.30 \, \cos 40^ ext{°}) \). Calculate these numerically.

Key Concepts

Kinetic Friction

Kinetic friction is a crucial concept when analyzing moving objects on surfaces. It occurs between the surfaces when an object is in motion, as opposed to static friction which acts when an object is at rest. In the context of the toboggan sliding on a snowy hill, kinetic friction comes into play as the toboggan moves both uphill and downhill.

- **Coefficient of Kinetic Friction**: This is a dimensionless value denoted as \( \mu_k \), which quantifies the degree of friction between two surfaces. For the given problem, \( \mu_k = 0.30 \), indicating moderate friction between toboggan and snow.
- **Kinetic Frictional Force**: This force can be calculated using the formula \( F_{f} = \mu_k F_{g, \perp} \). Here, \( F_{g, \perp} \) is the component of the gravitational force perpendicular to the surface of the incline. This frictional force is essential in determining the toboggan's acceleration.

Remember, the kinetic friction always acts opposite to the direction of motion, slowing down the sliding object.

Gravitational Force

Gravitational force is what pulls the toboggan towards the hill's surface. It's a constant force that can be split into two components when dealing with inclined planes.

- **Parallel Component (\( F_{g, \parallel} \))**: This part of the gravitational force acts along the slope and is calculated by \( mg \sin \theta \). It aids in moving the toboggan downwards.
- **Perpendicular Component (\( F_{g, \perp} \))**: This is the part of the gravitational force that acts perpendicular to the slope, given by \( mg \cos \theta \). This component is what the normal force counteracts.

By breaking down the gravitational force into these components, it's possible to more effectively analyze and solve problems involving motion on inclined planes.

Inclined Plane Dynamics

Understanding the dynamics of inclined planes is key to solving this type of physics problem. Inclines introduce a unique set of forces and require careful breakdown to resolve the motion.

- **Normal Force**: This is the support force exerted by a surface, responding to the perpendicular gravitational force component. On an incline, it is reduced to \( mg \cos \theta \).
- **Net Force on Incline**: Movement on an incline results from the balance of gravitational, frictional, and other forces. For upward and downward motion on a slope, the direction and magnitude of forces change.
- **Sum of Forces**: When solving for acceleration, sum the parallel component of gravity and kinetic friction. Whether the toboggan is going up or down alters the net force equation.

By fully understanding these forces, students can confidently tackle more advanced problems involving inclined terrains.

Acceleration Calculation

Calculating acceleration on an inclined plane involves understanding the net forces acting on the object.

1. **Up the Hill**: - Combine the downward gravitational force component and kinetic friction. Since friction and the incline gravity component work together, add these forces. - The formula \( a = g(\mu_k \cos \theta + \sin \theta) \) captures the net acceleration up the slope.

2. **Down the Hill**: - Here, the gravitational force aids the motion, while friction opposes it. Subtract the frictional force from the gravitational force parallel to the slope. - Use \( a = g(\sin \theta - \mu_k \cos \theta) \) to determine the net acceleration.

With given values \( (g = 9.8 \, \text{m/s}^2, \theta = 40^\circ) \), inserting them into the formulas will yield the precise accelerations for both upward and downward trajectories.