Problem 30 - College Algebra with Corequisite Support

Question

For the following exercises, use the formula given to solve for the required value. \(F=m a\) indicates that force \((F)\) equals mass \((m)\) times acceleration (a). Find the acceleration of a mass of \(50 \mathrm{~kg}\) if a force of \(12 \mathrm{~N}\) is exerted on it.

Step-by-Step Solution

Answer

The acceleration is 0.24 m/s².

Step 1: Understand the given formula

We are given the formula for force: \( F = m \times a \). This describes the relationship between force \( F \), mass \( m \), and acceleration \( a \). We need to solve for acceleration.

Step 2: Rearrange the formula to solve for acceleration

To find the acceleration, rearrange the formula \( F = m \times a \) to solve for \( a \). This results in \( a = \frac{F}{m} \).

Step 3: Substitute known values into the formula

Plug the given values into the equation \( a = \frac{F}{m} \). Here, \( F = 12 \text{ N} \) and \( m = 50 \text{ kg} \). So, the equation becomes \( a = \frac{12}{50} \).

Step 4: Calculate the value of acceleration

Perform the calculation: \( a = \frac{12}{50} = 0.24 \text{ m/s}^2 \). Thus, the acceleration is \( 0.24 \text{ m/s}^2 \).

Key Concepts

Force and Motion

Force and motion are closely intertwined concepts in physics that describe how objects move. Newton's Second Law, which is expressed as \( F = m \times a \), truly emphasizes this relationship. In this formula, \( F \) represents the force, \( m \) is the mass, and \( a \) denotes acceleration. This means that when you apply a force to an object, you are essentially trying to change how it moves.
Force can cause an object to start moving, stop moving, or change direction. Think of a soccer ball lying on the ground. It will remain still until a force, like a kick, causes it to roll or accelerate. The greater the force, the more motion or acceleration the object can experience.
Motion is not just about speed but also the change in velocity or direction over time. This is why force is a key factor in changing an object's motion, making it an important concept in all moving systems.

• Force affects acceleration and direction.
• Motion results from applied forces.

Understanding these concepts allows science students to predict how objects will behave in various scenarios, forming the foundation for much of physics.

Acceleration Calculation

To calculate acceleration, understanding the rearrangement of the formula \( F = m \times a \) is essential. When you want to find the acceleration, you'll rearrange it to \( a = \frac{F}{m} \). This calculation tells us how fast something speeds up when a force acts upon it.
In the exercise problem, we were given the force \( F = 12 \text{ N} \) and the mass \( m = 50 \text{ kg} \). Plugging these values into the equation, the acceleration \( a \) is calculated as \( a = \frac{12}{50} \).
Let's break it down:

• Start with the formula \( a = \frac{F}{m} \).
• Substitute \( 12 \text{ N} \) for \( F \) and \( 50 \text{ kg} \) for \( m \).
• Solve \( a = \frac{12}{50} = 0.24 \text{ m/s}^2 \).

So, the ball experiences an acceleration of \( 0.24 \text{ m/s}^2 \) which indicates how quickly it changes its velocity. Understanding and calculating acceleration helps us predict vehicular dynamics, sports mechanics, and various engineering problems involving motion.

Physics in Algebra

When exploring physics, algebra becomes a valuable tool for solving equations and understanding relationships between physical concepts. Newton's laws are a great example of where physics and algebraically solving neat formulas intersect.
In the calculation of acceleration from the formula \( F = m \times a \), algebra allows us to deftly rearrange and solve for a single variable. It involves basic algebraic manipulation to isolate \( a \), converting a physics law into something computationally useful.
This use of algebra in physics enables students to:

• Understand relationships between different physical quantities.
• Simplify complex scientific concepts into manageable problems.
• Apply mathematical tools to predict outcomes and validate experimental data.

The algebraic techniques used in physics strengthen problem-solving skills and support a deeper understanding of the material universe. Mastery in such areas not only helps in academic growth but also in real-world applications beyond classroom walls.