Problem 3 - Applied Physics

Question

Solve each formula for the quantity given. $$ w=m g \text { for } m $$

Step-by-Step Solution

Answer

The mass \( m \) is given by \( m = \frac{w}{g} \).

Step 1: Understand the Formula

The given formula is \( w = m g \), where \( w \) is the weight, \( m \) is the mass, and \( g \) is the acceleration due to gravity. Our goal is to solve this equation for the mass \( m \).

Step 2: Isolate the Variable 'm'

To solve for \( m \), we need to get \( m \) by itself on one side of the equation. We can do this by dividing both sides of the equation by \( g \).

Step 3: Divide Both Sides by 'g'

When we divide both sides by \( g \), the equation becomes \( \frac{w}{g} = m \). This isolates \( m \) on one side of the equation.

Step 4: Confirm the Solution

The final step is to write the isolated variable solution: \( m = \frac{w}{g} \). By confirming the solution, we have correctly solved for \( m \) according to the problem requirements.

Key Concepts

Weight and Mass Relationship

In physics, understanding the relationship between weight and mass is crucial. Weight and mass are related but different concepts.
Weight (\( w \)) is the force exerted by gravity on an object. It depends on both the mass of the object (\( m \)) and the gravitational acceleration (\( g \)). The formula forming this relationship is \( w = m g \). Here, \( g \) is approximately \( 9.8 \ m/s^2 \) on Earth.

Mass, on the other hand, is a measure of the amount of matter within an object. It remains the same regardless of location. Understanding these differences helps clarify why astronauts float in space—it's not that their mass changes, but the gravity affecting their weight changes.

• Weight is variable and depends on gravity.
• Mass is constant and intrinsic to the object.

Understanding the formula helps comprehend the broader physics concepts of force and acceleration.

Acceleration due to Gravity

The acceleration due to gravity is a fundamental concept in physics that explains the force acting on objects due to the Earth's gravitational pull. This force causes objects to fall towards the ground. On Earth, this acceleration is approximately \( 9.8 \ m/s^2 \), though it can vary slightly depending on geographic location and altitude.

This constant is crucial in calculating various physical properties, such as weight. When using the formula \( w = m g \), \( g \) acts as the conversion factor between mass and weight.

To visualize, think of it as the speed at which objects are pulled toward the earth. It influences all sorts of motion phenomena, from free-falling objects to the orbits of satellites.

• \( g \) is the same for all objects near Earth's surface.
• It provides a consistent way to calculate weight from mass.

Increasing your grasp of \( g \) deepens understanding of motions encountered in everyday life.

Algebraic Manipulation

Algebraic manipulation refers to rearranging equations to isolate different variables, making it easier to solve for unknowns. In the formula \( w = m g \), we can solve for \( m \) by isolating the variable on one side of the equation. This requires dividing both sides of the equation by \( g \), yielding \( m = \frac{w}{g} \).

This technique is not only useful in physics but also in numerous mathematical contexts.

• Isolating variables helps find solutions.
• It involves performing the same operation on both sides of an equation.

The ability to manipulate equations is fundamental in problem-solving across sciences and engineering disciplines. Through practice, one can easily rearrange complex formulas to find necessary solutions efficiently.