Foam Mass Calculation: Physics Problem Solved!

Hey guys! Let's dive into an interesting physics problem that involves calculating the mass of a piece of foam. This is a classic example that combines concepts of buoyancy, density, and equilibrium. We're going to break down the problem step by step, making it super easy to understand. So, let's get started!

Understanding the Buoyancy Basics

Before we jump into the calculations, let's quickly recap the fundamental principles at play here. The main concept is buoyancy, which is the upward force exerted by a fluid (in this case, water) that opposes the weight of an immersed object. This buoyant force is what makes objects float or at least seem lighter in water. The magnitude of this force is equal to the weight of the fluid displaced by the object, a principle famously stated by Archimedes. In simpler terms, if you dunk something in water, the water pushes back up with a force equal to the weight of the water that the object has moved out of the way. Now, let's add another layer: density. Density is simply how much "stuff" is packed into a certain space. We measure it as mass per unit volume (kg/m³). Objects with lower density than water tend to float, while denser objects sink. Think of a feather versus a rock; the rock is much denser and sinks like a charm, while the feather, being less dense, floats gracefully. So, when we talk about the foam in our problem, we know it's less dense than water (50 kg/m³ compared to water's 1000 kg/m³), which is why it floats in the first place! But here's the catch: to fully submerge it, we need an extra force – our student's weight. This brings us to the idea of equilibrium. Equilibrium is when all forces acting on an object are balanced, resulting in no net force and no acceleration. In our case, for the foam to be fully submerged and stay there, the downward forces (weight of the foam + weight of the student) must equal the upward buoyant force. Understanding these concepts of buoyancy, density, and equilibrium is crucial for solving this problem. So, let's keep these in mind as we move on to the next part where we'll dissect the problem statement and figure out exactly what we need to calculate. Remember, physics is all about understanding these core principles and applying them to real-world scenarios – or in this case, a cool foam-in-water scenario!

Dissecting the Problem Statement

Alright, let's break down this problem piece by piece. We've got a student, some foam, and a tub of water. The goal? To figure out the mass of the foam. The problem gives us a bunch of key information that we need to organize. First off, we know the student's mass is 38 kg. This is important because the student is adding weight to the foam to submerge it completely. Think of it like this: the foam wants to float, but the student is pushing it down. The fact that the student doesn't get their shoes wet tells us that they're not fully submerged themselves – they're just applying enough force to push the foam under. Next up, we have the density of the foam, which is 50 kg/m³. Remember, density tells us how much mass is packed into a certain volume. This low density is why the foam floats in the first place! If it were denser than water, it would just sink like a stone. Then, we're given the density of water, which is 1000 kg/m³. This is a crucial piece of information because it helps us calculate the buoyant force – the upward push of the water on the foam. Water's relatively high density is what gives it such a strong buoyant force, allowing boats and even large ships to float. The key here is that the buoyant force is equal to the weight of the water displaced by the submerged object. So, if we know the volume of the submerged foam, we can figure out the weight of the water it's displacing and thus the buoyant force. Now, let's think about what the problem is really asking. We need to find the mass of the foam. To do this, we'll need to use all the information we have: the student's mass, the densities of the foam and water, and the principle of equilibrium. The principle of equilibrium is super important here. It tells us that when the foam is fully submerged and the student is standing on it, the forces acting downwards (the weight of the foam and the weight of the student) must be equal to the force acting upwards (the buoyant force of the water). By equating these forces, we can set up an equation that allows us to solve for the unknown mass of the foam. This is the heart of the problem-solving strategy. Once we have this equation, it's just a matter of plugging in the numbers and doing the math. So, let's keep dissecting this problem and get ready to set up that equation!

Setting Up the Equations

Okay, now for the fun part – setting up the equations! This is where we translate our understanding of the physics principles and the problem statement into mathematical language. Remember, the key here is the equilibrium condition: the forces pushing down must equal the forces pushing up. Let's break down the forces involved. First, we have the downward forces. These consist of the weight of the foam and the weight of the student. Weight, as you might recall, is the force of gravity acting on an object's mass. We calculate weight using the formula: Weight = mass × acceleration due to gravity (W = mg). Here, 'm' is the mass, and 'g' is the acceleration due to gravity, which is approximately 9.8 m/s². So, the weight of the foam (W_foam) is its mass (m_foam) times 'g', and the weight of the student (W_student) is their mass (m_student) times 'g'. We know the student's mass (38 kg), so we can easily calculate their weight. The mass of the foam is what we're trying to find, so we'll keep that as a variable. Next, we have the upward force, which is the buoyant force (F_buoyant). As we discussed earlier, the buoyant force is equal to the weight of the water displaced by the submerged foam. To calculate this, we need to know the volume of the water displaced, which is the same as the volume of the foam (V_foam). The weight of the displaced water is then its volume times its density (ρ_water) times 'g'. So, F_buoyant = V_foam × ρ_water × g. Now, we have a bit of a challenge: we don't know the volume of the foam directly. But we do know its density (ρ_foam). And remember, density is mass divided by volume (ρ = m/V). So, we can rearrange this formula to express the volume of the foam in terms of its mass and density: V_foam = m_foam / ρ_foam. This is a crucial step because it allows us to relate the foam's volume to its mass, which is what we're trying to find. Now we can substitute this expression for V_foam into the buoyant force equation: F_buoyant = (m_foam / ρ_foam) × ρ_water × g. Phew! We're getting there. Now we have expressions for all the forces involved. Finally, we can write down the equilibrium equation: W_foam + W_student = F_buoyant. Substituting the expressions we derived earlier, we get: (m_foam × g) + (m_student × g) = (m_foam / ρ_foam) × ρ_water × g. This equation looks a bit intimidating, but don't worry! It's just a matter of plugging in the known values and solving for m_foam. Notice that 'g' appears in every term, which means we can divide both sides of the equation by 'g' to simplify things. This gives us: m_foam + m_student = (m_foam / ρ_foam) × ρ_water. We're now one step closer to the final answer. The next step is to plug in the numbers and do the algebra to isolate m_foam. So, let's move on to the calculations!

Plugging in the Numbers and Solving

Alright, guys, let's get down to the nitty-gritty and plug in the numbers into our equation. We've already done the hard work of setting up the equation, so this part should be relatively straightforward. Remember our simplified equilibrium equation? It looks like this: m_foam + m_student = (m_foam / ρ_foam) × ρ_water. Now, let's substitute the values we know. We have: * m_student = 38 kg (mass of the student) * ρ_foam = 50 kg/m³ (density of the foam) * ρ_water = 1000 kg/m³ (density of water) Plugging these values into our equation, we get: m_foam + 38 kg = (m_foam / 50 kg/m³) × 1000 kg/m³. Now, let's simplify this equation. First, we can simplify the right side by dividing 1000 kg/m³ by 50 kg/m³: m_foam + 38 kg = m_foam × 20. This looks much more manageable, right? Now, our goal is to isolate m_foam on one side of the equation. To do this, let's subtract m_foam from both sides: 38 kg = 20 × m_foam - m_foam. This simplifies to: 38 kg = 19 × m_foam. Finally, to solve for m_foam, we divide both sides by 19: m_foam = 38 kg / 19. And there you have it! m_foam = 2 kg. So, the mass of the foam is 2 kilograms. Isn't that cool? We started with a seemingly complex problem involving buoyancy, density, and equilibrium, but by breaking it down step by step and applying the right physics principles, we were able to arrive at a clear and concise answer. This is the power of physics – taking real-world scenarios and making sense of them using mathematical tools. Now, let's take a step back and recap the entire process, highlighting the key steps and concepts we used to solve this problem.

Recapping the Solution

Okay, let's do a quick recap of what we've done to solve this problem. We started with a scenario: a student standing on a piece of foam to fully submerge it in water. Our mission was to figure out the mass of the foam. To tackle this, we broke the problem down into manageable steps, using our understanding of physics principles. First, we reviewed the key concepts of buoyancy, density, and equilibrium. We talked about how buoyancy is the upward force exerted by a fluid, how density is mass per unit volume, and how equilibrium means that all forces acting on an object are balanced. These concepts are the foundation for understanding the problem. Next, we dissected the problem statement, identifying all the given information and what we were trying to find. We knew the student's mass, the densities of the foam and water, and we needed to find the mass of the foam. We recognized that the equilibrium condition was crucial: the downward forces (weight of the foam and student) must equal the upward force (buoyant force). Then, we set up the equations. This involved expressing the weight of the foam and student in terms of their masses and the acceleration due to gravity (g). We also expressed the buoyant force in terms of the volume of water displaced, which was related to the volume of the foam. We used the density formula to relate the volume of the foam to its mass and density. This allowed us to write the equilibrium equation in terms of the mass of the foam, which was our unknown. After setting up the equations, we plugged in the numbers and solved for the mass of the foam. This involved substituting the given values into the equilibrium equation, simplifying the equation, and isolating the unknown variable (m_foam). We found that the mass of the foam was 2 kg. Finally, we arrived at our answer: the mass of the foam is 2 kilograms. This problem is a great example of how physics can be used to solve real-world scenarios. By understanding the underlying principles and applying them systematically, we were able to break down a complex problem into manageable steps and arrive at a clear solution. So, the next time you see something floating in water, remember the concepts of buoyancy, density, and equilibrium – you might just be able to figure out some cool things about it!