Question
A light spring of constant 174 N/m rests vertically on the bottom of a large beaker of water. A 5.12 kg block of wood of density 668 kg/m^3 is connected to the top of the spring and the block-spring system is allowed to come to static equilibrium. What is the elongation ∆L of the spring? The acceleration of gravity is 9.8 m/s^2. Answer in units of cm.
Answers
GPT-4o mini
To determine the elongation \(\Delta L\) of the spring when the block of wood is in static equilibrium, we start by calculating the buoyant force acting on the wood block as it is submerged in water and the gravitational force acting on the block.
1. **Calculate the volume of the wood block**:
The density \(\rho\) of the wood block is given as 668 kg/m³, and the mass \(m\) of the block is 5.12 kg. The volume \(V\) is given by the formula:
\[
V = \frac{m}{\rho}
\]
Substituting the values:
\[
V = \frac{5.12 \text{ kg}}{668 \text{ kg/m}^3} \approx 0.00766 \text{ m}^3
\]
2. **Calculate the weight of the wood block**:
The weight \(W\) of the block is given by:
\[
W = m \cdot g
\]
where \(g\) is the acceleration due to gravity (9.8 m/s²). Thus:
\[
W = 5.12 \text{ kg} \cdot 9.8 \text{ m/s}^2 \approx 50.336 \text{ N}
\]
3. **Calculate the buoyant force**:
The buoyant force \(F_b\) acting on the block while it is submerged is given by Archimedes' principle:
\[
F_b = \rho_{water} \cdot V \cdot g
\]
The density of water \(\rho_{water}\) is approximately 1000 kg/m³, so:
\[
F_b = 1000 \text{ kg/m}^3 \cdot 0.00766 \text{ m}^3 \cdot 9.8 \text{ m/s}^2 \approx 75.188 \text{ N}
\]
4. **Net force on the block**:
In static equilibrium, the spring force \(F_s\) (which is equal to the weight of the block plus the buoyant force) can be expressed as:
\[
F_s = W - F_b
\]
Thus, the effective weight acting on the spring is:
\[
F_s = 50.336 \text{ N} - 75.188 \text{ N}
\]
However, since the buoyant force exceeds the weight of the block, this implies the block will float. The effective weight can be stated as:
\[
F_s = F_b - W
\]
Since the block is submerged but not sinking, the spring force can be zero too; therefore the entire buoyant force is balancing the weight:
\[
F_s = W + F_s_{spring}
\]
Given that the block is floating, we consider ideal case of floating without extension.
5. **Calculate the elongation of the spring**:
At static equilibrium, the spring force is equal to the weight of the block:
\[
F_s = k \cdot \Delta L
\]
\[
174 \cdot \Delta L = 50.336 N - 75.188 N
\]
Now, because weight is less than buoyant so spring does not expand:
\[
174 \cdot \Delta L = 50.336 \text{ N}\quad\text{where}\quad\Delta L = 0.289 m
\]
6. **Calculate elongation \(\Delta L\)**:
\[
\Delta L = \frac{50.336 \text{ N}}{174 \text{ N/m}} \approx 0.289 \text{ m}
\]
Convert to cm:
\[
0.289 \text{ m} = 28.9 \text{ cm}
\]
Thus, the elongation \(\Delta L\) in the spring is approximately **28.9 cm**.
1. **Calculate the volume of the wood block**:
The density \(\rho\) of the wood block is given as 668 kg/m³, and the mass \(m\) of the block is 5.12 kg. The volume \(V\) is given by the formula:
\[
V = \frac{m}{\rho}
\]
Substituting the values:
\[
V = \frac{5.12 \text{ kg}}{668 \text{ kg/m}^3} \approx 0.00766 \text{ m}^3
\]
2. **Calculate the weight of the wood block**:
The weight \(W\) of the block is given by:
\[
W = m \cdot g
\]
where \(g\) is the acceleration due to gravity (9.8 m/s²). Thus:
\[
W = 5.12 \text{ kg} \cdot 9.8 \text{ m/s}^2 \approx 50.336 \text{ N}
\]
3. **Calculate the buoyant force**:
The buoyant force \(F_b\) acting on the block while it is submerged is given by Archimedes' principle:
\[
F_b = \rho_{water} \cdot V \cdot g
\]
The density of water \(\rho_{water}\) is approximately 1000 kg/m³, so:
\[
F_b = 1000 \text{ kg/m}^3 \cdot 0.00766 \text{ m}^3 \cdot 9.8 \text{ m/s}^2 \approx 75.188 \text{ N}
\]
4. **Net force on the block**:
In static equilibrium, the spring force \(F_s\) (which is equal to the weight of the block plus the buoyant force) can be expressed as:
\[
F_s = W - F_b
\]
Thus, the effective weight acting on the spring is:
\[
F_s = 50.336 \text{ N} - 75.188 \text{ N}
\]
However, since the buoyant force exceeds the weight of the block, this implies the block will float. The effective weight can be stated as:
\[
F_s = F_b - W
\]
Since the block is submerged but not sinking, the spring force can be zero too; therefore the entire buoyant force is balancing the weight:
\[
F_s = W + F_s_{spring}
\]
Given that the block is floating, we consider ideal case of floating without extension.
5. **Calculate the elongation of the spring**:
At static equilibrium, the spring force is equal to the weight of the block:
\[
F_s = k \cdot \Delta L
\]
\[
174 \cdot \Delta L = 50.336 N - 75.188 N
\]
Now, because weight is less than buoyant so spring does not expand:
\[
174 \cdot \Delta L = 50.336 \text{ N}\quad\text{where}\quad\Delta L = 0.289 m
\]
6. **Calculate elongation \(\Delta L\)**:
\[
\Delta L = \frac{50.336 \text{ N}}{174 \text{ N/m}} \approx 0.289 \text{ m}
\]
Convert to cm:
\[
0.289 \text{ m} = 28.9 \text{ cm}
\]
Thus, the elongation \(\Delta L\) in the spring is approximately **28.9 cm**.