Problem 24 - College Physics

Question

\(\bullet\) \(\bullet\) A block of ice with mass 2.00 kg slides 0.750 m down an inclined plane that slopes downward at an angle of \(36.9^{\circ}\) below the horizontal. If the block of ice starts from rest, what is its final speed? You can ignore friction.

Step-by-Step Solution

Answer

The block's final speed is approximately 2.97 m/s.

Step 1: Identify the Given Information

We know the mass of the block of ice is 2.00 kg, the distance it slides down the ramp is 0.750 m, and the angle of the incline is \(36.9^{\circ}\). The block starts from rest, so its initial speed is 0 m/s. We need to find its final speed.

Step 2: Calculate the Acceleration Due to Gravity Along the Incline

The component of the gravitational force acting along the incline is calculated by \( F_{\text{gravity}} = m \, g \, \sin(\theta) \), where \( m \) is the mass (2.00 kg), \( g \) is the acceleration due to gravity (9.81 m/s²), and \( \theta \) is the angle of the incline (36.9°). Therefore, the acceleration along the incline is \( a = g \, \sin(\theta) = 9.81 \, \sin(36.9\degree) \approx 5.88 \, \text{m/s}^2 \).

Step 3: Apply the Kinematic Equation to Find Final Speed

Use the kinematic equation \( v^2 = u^2 + 2as \), where \( u \) is the initial speed (0 m/s), \( a \) is the acceleration (5.88 m/s²), and \( s \) is the distance (0.750 m). Plug these into the equation: \( v^2 = 0 + 2 \times 5.88 \times 0.750 \). Simplifying gives \( v^2 = 8.82 \), therefore, \( v = \sqrt{8.82} \approx 2.97 \, \text{m/s} \).

Step 4: Confirm the Solution

Double-check calculations to ensure no steps were skipped or errors were made: using the equation \( v = \sqrt{8.82} \), confirms that the final speed of the block is approximately \( 2.97 \, \text{m/s} \).

Key Concepts

Kinematic Equations

Kinematic equations are essential tools in physics for solving problems involving motion. These equations describe how objects move under the influence of various forces. When an object is moving with constant acceleration, we can use the kinematic equations to determine one or more aspects of its motion, such as speed, position, or time.

In the given exercise, we have a block sliding down an inclined plane. Since friction is neglected, the acceleration is constant, influenced only by gravity. This allows us to apply the kinematic equation:

• \( v^2 = u^2 + 2as \)
• where \( v \) stands for the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( s \) represents the distance traveled.

This equation helps to calculate the final speed of the block as it slides down the ramp, starting from rest. Here, the initial velocity \( u \) is 0, simplifying our calculations.

Acceleration Due to Gravity

Acceleration due to gravity is a constant that affects how objects fall under the earth's gravitational pull. It is approximately \(9.81 \text{ m/s}^2\). On an inclined plane, the gravitational force is split into components.

To find the components of gravitational force on the incline, we only consider the component parallel to the surface. This is because this component is responsible for the block's acceleration down the plane.

• The component parallel to the incline is given by \( F_{\text{gravity}} = m \, g \, \sin(\theta) \).
• Substitute the given values: mass \( m = 2.00 \text{ kg} \), \( g = 9.81 \text{ m/s}^2 \), and angle \( \theta = 36.9^\circ \).

When you plug these into the equation, the acceleration \( a \) comes out to about \( 5.88 \text{ m/s}^2 \) along the plane.

Components of Force

Understanding the components of force is critical when analyzing motion on an inclined plane. The force of gravity, which acts downward, can be broken into two parts:

• One parallel to the incline (\( m \, g \, \sin(\theta) \)).
• One perpendicular to the incline (\( m \, g \, \cos(\theta) \)).

For this exercise, the parallel component is our main concern. This is what dictates the block's motion as friction is ignored. This force is responsible for accelerating the block down the incline. The perpendicular component does not affect the block's sliding rate because it is counteracted by the normal force exerted by the surface of the incline.

Final Speed Calculation

Final speed calculation is the end goal of kinematics problems involving motion with constant acceleration. We aim to find how fast an object is moving after traveling a certain distance. In our exercise, we used the kinematic equation \( v^2 = u^2 + 2as \) to determine the block's final speed.

Since the block started from rest, its initial speed \( u \) is 0, simplifying our calculation:

• Plug in the acceleration \( a = 5.88 \text{ m/s}^2 \) and distance \( s = 0.750 \text{ m} \) into the equation.
• This results in: \( v^2 = 0 + 2 \times 5.88 \times 0.750 \).

Solving for \( v \), we get \( \sqrt{8.82} \approx 2.97 \text{ m/s} \) as the block's final speed after sliding down the incline.