Problem 46 - College Physics

Question

\(\bullet$$\bullet\) 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 tobogan are 0.40 and 0.30 , respectively, and the hill slopes upward at \(40.0^{\circ}\) above the horizontal. Find the acceleration of the tobogan (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

The acceleration is approximately \(-3.50 \, \text{m/s}^2\) ascending and \(8.18 \, \text{m/s}^2\) descending.

Step 1: Understanding the Forces

To find the acceleration of the toboggan as it goes up and down the hill, we first need to identify all the forces acting on it. The forces include gravity, kinetic friction, and the normal force. We will use these forces to calculate the net force and, consequently, the acceleration.

Step 2: Calculating Forces for Ascending

While ascending, the forces acting on the toboggan are the gravitational force component along the hill, the kinetic friction force, and the normal force. The gravitational force component along the slope is calculated as:\[ F_{g, parallel} = mg \sin \theta \]The normal force \( F_N \) is:\[ F_N = mg \cos \theta \]The kinetic friction force \( F_k \) is:\[ F_k = \mu_k F_N = 0.3 \times mg \cos \theta \]

Step 3: Determining Net Force & Acceleration for Ascending

The net force acting on the toboggan as it ascends is the gravitational force minus the friction force. \[ F_{net, ascending} = mg \sin \theta - 0.3 \times mg \cos \theta \]Thus, the acceleration \( a_{up} \) is given by:\[ a_{up} = \frac{F_{net, ascending}}{m} = g \sin \theta - 0.3g \cos \theta \]Substitute known values and solve:\[ a_{up} = 9.8 \times (\sin 40^{\circ} - 0.3 \times \cos 40^{\circ}) \]

Step 4: Calculations for Descending Forces

For descending, the normal force and the gravitational force component have the same values as before. However, now, the friction force acts in the same direction as the motion, aiding the toboggan's acceleration downhill.Thus, the net force is:\[ F_{net, descending} = mg \sin \theta + 0.3 \times mg \cos \theta \]

Step 5: Determining Net Force & Acceleration for Descending

The acceleration \( a_{down} \) as the toboggan descends can be calculated by:\[ a_{down} = \frac{F_{net, descending}}{m} = g \sin \theta + 0.3g \cos \theta \]Substitute the values:\[ a_{down} = 9.8 \times (\sin 40^{\circ} + 0.3 \times \cos 40^{\circ}) \]

Step 6: Final Calculation and Result Summarization

Using trigonometric values, solve for both accelerations:- Going up: \( a_{up} = 9.8 \times (0.6428 - 0.3 \times 0.7660) \approx 3.50 \, \text{m/s}^2 \)- Going down: \( a_{down} = 9.8 \times (0.6428 + 0.3 \times 0.7660) \approx 8.18 \, \text{m/s}^2 \)

Key Concepts

Kinetic Friction

Kinetic friction is the force that opposes the motion of two surfaces sliding past each other. In our toboggan problem, kinetic friction comes into play when the toboggan is in motion up or down the hill.
The kinetic friction force can be calculated using the formula:

• \[ F_k = \, \mu_k \, F_N \]

where \( \mu_k \) is the coefficient of kinetic friction (0.30 in this case), and \( F_N \) is the normal force. This force acts parallel to the surface and opposes the direction of motion, slowing down the toboggan.
Understanding that kinetic friction is lower than static friction explains why objects in motion tend to stay in motion, once moved. For calculation, it is essential to use the kinetic frictional force while the toboggan is moving, whether uphill or downhill.

Static Friction

Static friction acts when an object is at rest relative to a surface, preventing it from sliding. For our toboggan scenario, although the problem primarily deals with motion, the static friction coefficient (\( \mu_s = 0.40 \) in this case) tells us how much grip the snow surface has.
In most real-world situations, static friction is what must be overcome to start the motion of an object. To calculate it, we would use:

• \[ F_s = \mu_s \, F_N \]

before the toboggan starts moving.
Static friction is generally higher than kinetic friction, which is why it's often harder to start moving an object than to keep it moving once it's in motion. This knowledge helps appreciate the initial forces at work in stopping or starting motion on inclined surfaces.

Inclined Plane

An inclined plane is a flat, sloped surface. In our problem, the toboggan moves on a hill sloping at 40 degrees. This inclination affects how forces such as gravity and friction act on the object.
The gravity force acting on an object is divided into two components:

• One parallel to the plane (\( F_{g, parallel} = mg \sin \theta \)), which tries to pull the toboggan downhill,
• And another perpendicular to the plane (\( F_N = mg \cos \theta \)), which affects the normal force.

To determine acceleration, these components are used in force calculations to resolve the net force on the toboggan as it moves up or down the slope.
Understanding the distribution of these gravitational forces is crucial for analyzing motion on inclined surfaces and predicting how much force is required to move objects across or along them.

Newton's Second Law

Newton's Second Law is a fundamental principle that describes how forces affect the motion of an object. It states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass, formulated as:

• \[ F_{net} = ma \]

In the toboggan problem, we apply this law to determine the acceleration when the net force is known. On the inclined plane, different forces—gravity, friction, and the normal force—combine to determine the net force.
The net force calculation changes depending on the direction of motion, whether ascending or descending the slope.

• Ascending, the net force is calculated by subtracting the kinetic friction force from the gravitational component, resulting in negative acceleration as the toboggan goes upwards.
• Descending, the net force adds the kinetic friction force to the gravitational component, leading to positive acceleration.

These calculations enable us to determine the precise accelerations, illustrating how Newton’s second law is used to understand motion dynamics on inclined planes.