Problem 12 - University Physics with Modern Physics

Question

A crate with mass 32.5 kg initially at rest on a warehouse floor is acted on by a net horizontal force of 14.0 N. (a) What acceleration is produced? (b) How far does the crate travel in 10.0 s? (c) What is its speed at the end of 10.0 s?

Step-by-Step Solution

Answer

(a) 0.431 m/s², (b) 21.55 m, (c) 4.31 m/s.

Step 1: Identify the Given Data

To solve the problem, we first need to identify the given values:- Mass of the crate \( m = 32.5 \text{ kg} \).- Net horizontal force \( F = 14.0 \text{ N} \).- Initial velocity \( u = 0 \text{ m/s} \) (since the crate is initially at rest).- Time \( t = 10.0 \text{ s} \).

Step 2: Calculate the Acceleration

Use Newton's second law to find the acceleration:\[ F = m \cdot a \]Where \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration. Rearrange the formula to solve for \( a \):\[ a = \frac{F}{m} = \frac{14.0}{32.5} \text{ m/s}^2 \approx 0.431 \text{ m/s}^2 \].

Step 3: Calculate the Distance Travelled

Use the kinematic equation for distance:\[ s = ut + \frac{1}{2} a t^2 \]Since the initial velocity \( u = 0 \), the equation simplifies to:\[ s = \frac{1}{2} \times 0.431 \times (10.0)^2 \]Calculate:\[ s = \frac{1}{2} \times 0.431 \times 100 = 21.55 \text{ m} \].

Step 4: Calculate the Final Speed

Use the kinematic equation for the final velocity:\[ v = u + at \]Since \( u = 0 \):\[ v = 0 + 0.431 \times 10.0 \]Calculate:\[ v = 4.31 \text{ m/s} \].

Key Concepts

Kinematic Equations

Kinematic equations play a crucial role when analyzing the motion of objects under the influence of constant forces. These equations allow us to predict an object's movement through parameters such as initial velocity, final velocity, acceleration, time, and displacement.

The basic set of kinematic equations include:

• Final velocity: \( v = u + at \)
• Displacement: \( s = ut + \frac{1}{2} a t^2 \)
• Final velocity squared: \( v^2 = u^2 + 2as \)

In these formulas, \( u \) is the initial velocity, \( v \) is the final velocity, \( a \) is the acceleration, \( t \) accounts for the time, and \( s \) represents displacement.

For example, if you know the object's acceleration and initial velocity, you can calculate how far it will travel over a given period by simply substituting values into the displacement equation. In the problem, since the crate was initially at rest, using the simplification for initial velocity (\( u = 0 \)) showed how to directly compute the distance with just acceleration and time values.

Acceleration Calculation

Determining an object's acceleration becomes intuitive when using Newton's Second Law of motion. This law states that the acceleration \( a \) of an object is directly proportional to the net force \( F \) acting upon it and inversely proportional to its mass \( m \). The mathematical expression for this law is:

• \( a = \frac{F}{m} \)

In our example, since the force applied to the crate is 14.0 N, and its mass is 32.5 kg, substituting these values into Newton’s formula gives:
\( a = \frac{14.0}{32.5} \approx 0.431 \text{ m/s}^2 \).

This acceleration means for every second, the crate's velocity changes by approximately 0.431 meters per second. Understanding the relationship between force, mass, and acceleration allows us to predict how objects will behave when forces are applied to them.

Force and Motion

The interaction between force and motion can be effectively understood by exploring Newton's Second Law. Newton's Second Law provides a framework to analyze how a force will change an object's motion.

When a net force is exerted on an object, it causes the object to accelerate in the direction of the force, proportional to the magnitude of the net force and inversely proportional to the object's mass. In simpler terms, heavier objects require more force to achieve the same acceleration as lighter ones.

Let's consider the crate example: a net horizontal force of 14.0 N causes it to accelerate at approximately 0.431 m/s². This change in motion over time results in two important outcomes:

• The crate's position changes (it moves a distance, calculated using kinematic equations).
• The crate's speed increases (final speed can be determined using its acceleration and the time factor).

This understanding helps visualize how varied forces alter motions, affecting factors such as distance traveled and speed achieved over time. It brings to life the intricate link between force and the resulting motion, as demonstrated with our example of the crate on a warehouse floor.