Answer :
At equilibrium, the block experiences a downward weight, countered by the upward buoyant force and spring force. The block undergoes an elongation (∆L) of approximately -0.2108 meters, indicating a compression in the spring due to the balance of forces in the system.
1. Understand the Forces Involved:
Weight of the block (downward): This is calculated as the mass of the block times the acceleration due to gravity (Weight = m * g).
Buoyant force (upward): This is the force exerted by the water on the submerged block. It's equal to the weight of the water displaced by the block (Buoyant Force = density of water * volume of block submerged * g).
Spring force (upward): This force is determined by Hooke's Law (Spring Force = spring constant * elongation).
2. Analyze the Situation at Equilibrium:
When the block-spring system is in equilibrium, the net force acting on the block is zero. This means the weight of the block is balanced by the sum of the buoyant force and the spring force.
3. Set Up Equations:
Weight of the block: Weight = m * g
= 5.64 kg * 9.8 m/s²
= 55.272 N
Volume of the block: Volume = mass / density
= 5.64 kg / 604 kg/m³
= 0.00934 m³
Buoyant force: Buoyant force = density of water * Volume * g
= 1000 kg/m³ * 0.00934 m³ * 9.8 m/s²
= 91.532 N
4. Apply Equilibrium Condition:
Weight of block = Buoyant force + Spring force
Substitute the values from step 3: 55.272 N
= 91.532 N + 172 N/m * ∆L
5. Solve for the Elongation (∆L):
Rearrange the equation to isolate ∆L: ∆L
= (55.272 N - 91.532 N) / 172 N/m
Calculate the result: ∆L = -0.2108 m
Complete question:
A light spring of constant 172 N/m rests vertically on the bottom of a large beaker of water. A 5.64 kg block of wood of density 604 kg/m³ 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/s².
