Problem 94 - Physics for Scientists and Engineers with Modern Physics

Question

A 16 -kg sled starts up a \(28^{\circ}\) incline with a speed of \(2.4 \mathrm{~m} / \mathrm{s}\). The coefficient of kinetic friction is \(\mu_{k}=0.25 .\) (a) How far up the incline does the sled travel? ( \(b\) ) What condition must you put on the coefficient of static friction if the sled is not to get stuck at the point determined in part \((a) ?\) (c) If the sled slides back down, what is its speed when it returns to its starting point?

Step-by-Step Solution

Answer

(a) The sled travels approximately 0.53 meters up the incline. (b) The coefficient of static friction must satisfy \( \mu_s > \tan \theta - \mu_k \cos \theta \). (c) The sled returns with a speed of 2.4 m/s.

Step 1: Analyzing Forces Involved

The forces acting on the sled include gravity, which acts downward, and friction, which opposes motion up the slope. The component of gravitational force acting down the incline is \( mg \sin \theta \) and the normal force is \( mg \cos \theta \). The force of friction is \( f_k = \mu_k \cdot N = \mu_k \cdot mg \cos \theta \).

Step 2: Applying Newton's Second Law

The net force acting on the sled as it moves up the incline is the sum of gravity and friction in the opposite direction of the initial velocity: \( F_{net} = -mg \sin \theta - \mu_k mg \cos \theta \). By Newton's second law, \( F = ma \), the acceleration \( a \) is thus: \( a = \frac{-mg \sin \theta - \mu_k mg \cos \theta}{m} = -g \sin \theta - \mu_k g \cos \theta \).

Step 3: Using Kinematics to Find Distance

We use the kinematic equation: \( v^2 = u^2 + 2a s \) to find the distance \( s \) where the final velocity \( v = 0 \), the initial velocity \( u = 2.4 \text{ m/s} \), and \( a \) is as calculated. Rearrange this to solve for \( s \): \( 0 = (2.4)^2 + 2(-g \sin \theta - \mu_k g \cos \theta)s \). Solve for \( s \): \( s = \frac{(2.4)^2}{2(g \sin \theta + \mu_k g \cos \theta)} \).

Step 4: Calculating the Critical Coefficient of Static Friction

To ensure the sled doesn't get stuck, the static friction force must be less than or equal to the forces pulling it down: \( f_s = \mu_s \cdot N \leq mg \sin \theta + \mu_k mg \cos \theta \). Solve for \( \mu_s \): \( \mu_s = \frac{mg \sin \theta + \mu_k mg \cos \theta}{mg \cos \theta} \).

Step 5: Determine Speed on Return Journey

On returning, the sled starts with \( v = 0 \), acceleration due to gravity and friction: \( a = g \sin \theta - \mu_k g \cos \theta \). Use the kinematic equation again: \( v^2 = 0 + 2(g \sin \theta - \mu_k g \cos \theta)s \) to find speed \( v \) at starting point after sliding back, using \( s \) found in Step 3.

Key Concepts

Kinematic Equations

Kinematic equations are fundamental in physics, particularly when studying motion. These equations relate various variables such as initial velocity, final velocity, acceleration, time, and displacement. In our sled problem, we used the kinematic equation for motion:

• \( v^2 = u^2 + 2a s \)

This formula helps us find the distance \( s \) the sled travels up the incline. Here:

• \( v \) is the final velocity,
• \( u \) is the initial velocity,
• \( a \) is acceleration, and
• \( s \) is the displacement.

Since the final velocity is zero when the sled stops, we rearrange the equation to solve for \( s \), allowing us to find how far the sled travels before it stops. Understanding these equations allows us to predict how objects move under various forces.

Newton's Second Law

Newton's second law is a core principle in mechanics that explains how forces affect motion. It states that the force acting on an object is equal to the mass of the object multiplied by its acceleration:

• \( F = ma \)

In our scenario, several forces act on the sled as it moves up the incline:

• Gravitational force down the slope: \( mg \sin \theta \)
• Normal force perpendicular to the slope: \( mg \cos \theta \)
• Frictional force opposing motion: \( f_k = \mu_k \cdot mg \cos \theta \)

By applying Newton's second law, we find the net force by adding these up, and then divide by mass \( m \) to find the sled's acceleration \( a \). This calculation gives insight into how forces slow the sled as it ascends the incline.

Friction

Friction plays a significant role in the motion of objects on surfaces. It’s a force that opposes motion between two surfaces that are in contact.

• Kinetic friction: occurs when objects are sliding over each other.
• Static friction: prevents surfaces from starting to slide.

In our sled problem, kinetic friction is calculated using:

• \( f_k = \mu_k \cdot mg \cos \theta \)

where \( \mu_k \) is the coefficient of kinetic friction. It takes into account the nature of the surfaces. Static friction becomes crucial when considering if the sled might get stuck after stopping. By calculating a critical static friction value, we ensure that it can start moving again without getting stuck halfway.

Inclined Plane Motion

Inclined plane motion involves movement along a sloped surface, which introduces unique forces compared to flat surface motion. The slope impacts how gravitational force is divided into components:

• Parallel component (causes sliding down): \( mg \sin \theta \)
• Perpendicular component (acts as normal force): \( mg \cos \theta \)

When dealing with these problems, it's essential to consider how both gravity and friction interact on the slope. These components affect how fast or slow an object moves. In our exercise, understanding these forces allows us to determine how far the sled travels uphill before stopping, and how to ensure it starts moving again if it doesn’t get stuck due to static friction. These concepts provide clarity on the mechanics of motion on inclined surfaces, combining forces in ways that flat surfaces don't present.