Respuesta :
Answer: 24.5 cm
Explanation:
Given
Force constant of spring, k = 40 N/m
Diameter of spring, d = 5.1 cm = 0.051 m
Mass of cylinder, m = 1.5 kg
Let us assume that the cylinder is hanging in such a way that the circular end is parallel with the water. Also, we assume that the tank water level is not materially affected by the displacement of the cylinder while the cylinder sinks. The water is fresh and as we all know, the density of water is 1000 kg/m³
To solve this, we assume x to be the spring extension and it's equivalent sinking distance(in meters). We then apply the formula,
mg = kx + ρgAx
mg = x(k + ρgA)
x = mg / (k + ρgA), where
A = πd²/4
A = (3.142 * 0.051²)/4
A = 0.0082 / 4
A = 0.00205 m²
x = 1.5 * 9.81 / [40 + (1000 * 9.81 * 0.00205)]
x = 14.715 / (40 + 20.1105)
x = 14.715 / 60.1105
x = 0.245 m
or 24.5 cm of stretch or sinking
When in equilibrium, the length of the cylinder that is submerged is; x = 24.5 cm
We are given;
Spring Constant; k = 40 N/m
Diameter of spring; d = 5.1 cm = 0.051 m
Mass of the metal cylinder; m = 1.5 kg
We are told that when released the cylinder will reach an equilibrium position. This means it will extend by a length of say "x".
To find x the length of the cylinder that was submerged, we will use the formula at equilibrium position which is;
mg = kx + ρgAx
Where;
m is mass of cylinder
g is acceleration due to gravity = 9.81 m/s²
k is spring constant
ρ is density of water = 1000 kg/m³
A is area of spring = πd²/4 = π(0.051²)/4
x is length by which cylinder was submerged.
Thus;
1.5 × 9.81 = x(40 + (1000 × 9.81 × π(0.051²)/4)
14.715 = x(40 + 20.04)
x = 14.715/60.04
x = 0.245 m or 24.5 cm
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