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Chapter 9: Problem 78

A cube that is \(4.00 \mathrm{cm}\) on a side and of density $8.00 \times 10^{2} \mathrm{kg} / \mathrm{m}^{3}$ is attached to one end of a spring. The other end of the spring is attached to the base of a beaker. When the beaker is filled with water until the entire cube is submerged, the spring is stretched by \(1.00 \mathrm{cm} .\) What is the spring constant?

Short Answer

Expert verified
Answer: The spring constant of the spring is 439.2 N/m.

Step by step solution

01

Calculate the mass of the cube

We are given the density (\(\rho\)) of the cube and its side length (\(a\)). To find the mass (\(m\)) of the cube, we can use the formula \(m=\rho V\), where \(V\) is the volume of the cube and can be calculated as \(a^3\). The volume of the cube is: \(V = a^3 = (4.00\,\text{cm})^3 = 64.00\,\text{cm}^3.\) Since we're dealing with density in kg/m鲁, we must convert the volume to m鲁: \(V = 64.00 \times 10^{-6}\,\text{m}^3\) Now we can calculate the mass of the cube: \(m = \rho V = (8.00 \times 10^2\,\text{kg/m}^3) \times (64.00 \times 10^{-6}\,\text{m}^3) = 0.512\,\text{kg}.\)
02

Calculate the weight of the cube

To find the weight (\(W\)) of the cube, we can use the formula \(W = m g\), where \(g\) is the acceleration due to gravity (approximately \(9.81\,\text{m/s}^2\)). The weight of the cube is: \(W = mg = (0.512\,\text{kg}) \times (9.81\,\text{m/s}^2) = 5.019\,\text{N}\).
03

Calculate the buoyant force exerted by the fluid

The buoyant force (\(F_B\)) exerted by the fluid on the submerged object (cube) is equal to the weight of the displaced fluid. We can calculate the buoyant force using the formula \(F_B = \rho_{water} V g\), where \(\rho_{water}\) is the density of water (approximately \(1 \times 10^3\,\text{kg/m}^3\)). The buoyant force exerted by the water is: \(F_B = \rho_{water} V g = (1 \times 10^3\,\text{kg/m}^3) \times (64.00 \times 10^{-6}\,\text{m}^3) \times (9.81\,\text{m/s}^2) = 0.627\,\text{N}\).
04

Use Hooke's law to find the spring constant

According to Hooke's law, the force exerted by the spring (\(F_s\)) is proportional to its displacement from the equilibrium position (\(x\)). The force exerted by the spring balances the difference between the weight of the cube and the buoyant force, so \(F_s = W - F_B\). The force exerted by the spring is: \(F_s = W - F_B = 5.019\,\text{N} - 0.627\,\text{N} = 4.392\,\text{N}\). Now, we can use the formula \(F_s = kx\) to solve for the spring constant (\(k\)), where \(x = 1.00\,\text{cm} = 0.0100\,\text{m}\) The spring constant is: \(k = \frac{F_s}{x} = \frac{4.392\,\text{N}}{0.0100\,\text{m}} = 439.2\,\text{N/m}\). The spring constant of the spring is \(439.2\,\text{N/m}\).

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