A spiral spring, loaded with a piece os metal, extends by 10.5 cm in air, when the metal is fully submerged in water, the spring extends by 6.8 cm. Calculate the relative density of the metal. ( Assuming Hooke’s law is obeyed).
Fwater = k x = Volume * (density in water) * 6.8
Fair = k x = Volume * (density in air) * 10.5
density in air /density in water =10.5/6.8 = 1.54
To calculate the relative density of the metal, we can use the concept of buoyancy and Hooke's law.
Let's assume that the original length of the spring (when no load is applied) is denoted by L.
In air:
Extension = 10.5 cm
In water:
Extension = 6.8 cm
Now, let's consider the forces acting on the metal and the spring in both cases.
In air:
The force exerted by the spring is proportional to the extension according to Hooke's law. Thus, we can write:
Force air = k * extension air
In water:
The force exerted by the spring is again proportional to the extension according to Hooke's law, but there is an additional force due to buoyancy. So, we can write:
Force water = k * extension water + Buoyant force
Since the volume of the metal is the same in both cases, the increase in volume when submerged in water is equal to the volume of water displaced by the metal. According to Archimedes' principle, the buoyant force is equal to the weight of the water displaced. Therefore, it can be written as:
Buoyant force = Weight of the water displaced = mg
Now, let's equate the forces in air and water:
k * extension air = k * extension water + mg
We can rearrange the equation to solve for the mass m:
mg = k * (extension air - extension water)
m = k * (extension air - extension water) / g
The relative density (ρ) of the metal can be defined as the ratio of its density to the density of water. Therefore:
ρ = (density of metal) / (density of water)
We know that density = mass / volume. Rearranging, we get:
mass = density * volume
Since the density and volume of the metal are the same in both cases, we get:
density * volume = density * volume
Therefore, the relative density can be written as:
ρ = m / (volume of metal) = m / V
Now, substituting the earlier expression for mass m, we have:
ρ = (k * (extension air - extension water) / g) / V
Finally, substituting the known values:
ρ = (k * (10.5 cm - 6.8 cm) / g) / V
Now, we need to know the value of the spring constant k and the volume of the metal (V) to calculate the relative density of the metal.
To solve this problem, we need to apply Hooke's Law, which states that the force exerted by a spring is directly proportional to its extension. Mathematically, this can be written as:
F = kx
Where:
F = Force applied on the spring (in Newtons)
k = Spring constant (in Newtons per meter)
x = Extension of the spring (in meters)
In this case, we are given two scenarios: when the metal is in air and when it is fully submerged in water. Let's denote the two scenarios as follows:
1) Air scenario: Extension = 10.5 cm = 0.105 meters
2) Water scenario: Extension = 6.8 cm = 0.068 meters
Since the same metal is used in both scenarios, the force applied on the spring can be considered the same. Therefore, we can set up the following equation:
k * 0.105 = k * 0.068
The spring constant (k) cancels out, which leaves us with:
0.105 = 0.068
Now, let's rearrange the equation to solve for the relative density of the metal (ρ). The relative density of a substance can be defined as the ratio of its density (ρ₁) to the density of the reference substance (ρ₀), usually water.
ρ = ρ₁ / ρ₀
In this case, the reference substance is water, so ρ₀ = density of water = 1000 kg/m³.
Since the force applied on the spring is equal in both scenarios (air and water), and the extension is proportional to the force, we can write:
x₁ / x₀ = ρ₁ / ρ₀
Substituting the given extensions, we have:
0.105 / 0.068 = ρ₁ / 1000
To find the relative density of the metal (ρ₁), we rearrange the equation as follows:
ρ₁ = (0.105 / 0.068) * 1000
Simplifying the calculation:
ρ₁ = 1544.12
Therefore, the relative density of the metal is approximately 1544.12.