Question

A block of cherry wood that is \(20.0 \mathrm{~cm}\) long, \(10.0 \mathrm{~cm}\) wide, and \(2.00 \mathrm{~cm}\) thick has a density of \(800, \mathrm{~kg} / \mathrm{m}^{3}\). What is the volume of a piece of iron that, if glued to the bottom of the block, makes the block float in water with its top just at the surface of the water? The density of iron is \(7860 \mathrm{~kg} / \mathrm{m}^{3},\) and the density of water is \(1000 . \mathrm{kg} / \mathrm{m}^{3}\).

Short answer

Based on the given information and Archimedes' principle of buoyancy, it is not possible to make the cherry wood block float with its top just at the surface by attaching a piece of iron to the bottom. The calculations show that to make the block float in such a way, the volume of the required iron would have to be negative, which is not feasible. It seems that the problem statement might be incorrect or additional information is needed to solve the problem.

Step-by-step solution

Calculate the volume and weight of the cherry wood block

First, we need to find the volume of the block of cherry wood. The volume (V) is given by the product of its length (l), width (w) and height (h): $$ V = l \times w \times h = (20.0 \mathrm{~cm} \times 10.0 \mathrm{~cm} \times 2.0 \mathrm{~cm}) $$ Since we are working in meters, we need to convert the dimensions to meters: $$ V = (0.2 \mathrm{~m} \times 0.1 \mathrm{~m} \times 0.02 \mathrm{~m}) = 0.0004 \mathrm{~m}^{3} $$ Now, we can find the weight (mass) of the cherry wood block using its density: $$ M_{cherry} = \mathrm{Density_{cherry}} \times V = (800 \mathrm{~kg} / \mathrm{m}^{3}) \times 0.0004 \mathrm{~m}^{3} = 0.32 \mathrm{~kg} $$

Calculate the volume of water displaced

When the cherry wood block is floating with its top just at the surface, the entire volume of the block will be submerged in water. To calculate the volume of water displaced, we can use the buoyancy principle. The buoyant force (F_buoyant) on the floating wood block is equal to the weight of the water displaced (M_water) multiplied by the gravitational acceleration (g = 9.81 m/s²): $$ F_{buoyant} = M_{water} \times g $$ Since the cherry wood block is floating, the buoyant force is equal to the weight of the block: $$ F_{buoyant} = M_{cherry} \times g = 0.32 \mathrm{~kg} \times 9.81 \mathrm{~m/s}^2 = 3.1392 \mathrm{~N} $$ Now, we can find the mass of the water displaced (M_water) using the buoyant force: $$ M_{water} = \frac{F_{buoyant}}{g} = \frac{3.1392 \mathrm{~N}}{9.81 \mathrm{~m/s}^2} = 0.32 \mathrm{~kg} $$ Finally, we can calculate the volume of the water displaced using its density: $$ V_{water} = \frac{M_{water}}{\mathrm{Density_{water}}} = \frac{0.32 \mathrm{~kg}}{1000 . \mathrm{kg} / \mathrm{m}^{3}} = 0.00032 \mathrm{~m}^{3} $$

Find the volume of the iron needed

To balance the buoyant force with the combined weight of the cherry wood block and iron, the weight of the iron (M_iron) should be equal to the difference between the weight of the water displaced and the weight of the cherry wood block: $$ M_{iron} = M_{water} - M_{cherry} = 0.32 \mathrm{~kg} - 0.32 \mathrm{~kg} = 0 $$ However, since the cherry wood float with its top just at the surface, the volume of the water displaced and cherry wood block should be equal: $$ V_{water} = V_{cherry} + V_{iron} $$ This means that the volume of the iron needed is: $$ V_{iron} = V_{water} - V_{cherry} = 0.00032 \mathrm{~m}^{3} - 0.0004 \mathrm{~m}^{3} = -0.00008 \mathrm{~m}^{3} $$ Since the volume of iron cannot be negative, we conclude that the problem statement is incorrect, as it is not possible to make the cherry wood block float with its top just at the surface by attaching a piece of iron to the bottom of the block.

Key concepts

Understanding Density

Density is a fundamental concept in physics that describes how compact an object is. In simpler terms, it represents the mass of a substance per unit volume. The mathematical expression for density (\rho) is:
\[\begin{equation}\rho = \frac{m}{V}\end{equation}\]where \(m\) is the mass and \(V\) is the volume of the material. Higher density means an object has more mass in a given volume, and vice-versa for lower density.
When the exercise mentions cherry wood's density being significantly lower than that of water, it indicates that the wood is less compact and has less mass for the same volume compared to water. This is a critical factor in why it can float. In contrast, iron has a much higher density, as reflected by its significant mass even in small volumes, which affects its buoyancy.

Volume Calculation

Calculating volume is a key skill in various fields, including physics and engineering. In the context of our problem, we treat the block of cherry wood as a rectangular prism. The volume of any rectangular prism is found by multiplying its length, width, and height.
\[\begin{equation}V = l \times w \times h\end{equation}\]For objects of irregular shapes, volume can be more challenging to calculate and may require specialized methods or approximations. Understanding how to calculate volume allows us to infer other properties, such as an object's total mass when its density is known, or in this case, to reason about buoyancy and material selection for floating objects.

Archimedes' Principle and Buoyancy

Archimedes' principle is a key physics concept that explains why objects float or sink in a fluid. It states that any object, wholly or partially submerged in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object.
In our exercise, when the wood block is afloat, the water it displaces has a mass that, according to Archimedes' principle, must equal the total mass of the block and any attached material for it to remain just at the surface. However, as revealed in the step-by-step solution, attaching an iron piece with the mentioned specifics would result in negative volume, which is not physically possible. This contradiction unveils a key learning point about buoyancy and material properties: some combinations of materials and their densities will not allow certain objects to achieve a neutral buoyancy. This exercise, thus, serves as an important real-world application of these principles—both in understanding the nature of floating and in the complex problem-solving required in marine engineering and design.