Fish are hung on a spring scale to determine their mass. The spring stretches 5.00 cm for a load of 11.0 kg of fish. What mass of fish will stretch the spring 5.48 cm?
The correct answer and explanation is:
To solve this problem, we use the principle of Hooke’s Law, which states that the extension (stretch) of a spring is directly proportional to the applied force (as long as the elastic limit is not exceeded). Mathematically, Hooke’s Law is: F=kxF = kx
Where:
- FF is the force applied (in newtons, N)
- kk is the spring constant (in N/m)
- xx is the stretch or extension of the spring (in meters)
Step 1: Calculate the spring constant kk
First, convert 5.00 cm to meters: 5.00 cm=0.0500 m5.00 \text{ cm} = 0.0500 \text{ m}
The weight of 11.0 kg of fish is the force F=mgF = mg, where g=9.8 m/s2g = 9.8 \, \text{m/s}^2: F=11.0 kg×9.8 m/s2=107.8 NF = 11.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 107.8 \, \text{N}
Now calculate kk: k=Fx=107.8 N0.0500 m=2156 N/mk = \frac{F}{x} = \frac{107.8 \, \text{N}}{0.0500 \, \text{m}} = 2156 \, \text{N/m}
Step 2: Use kk to find the mass for 5.48 cm stretch
Convert 5.48 cm to meters: 5.48 cm=0.0548 m5.48 \text{ cm} = 0.0548 \text{ m}
Now find the force that causes this stretch: F=kx=2156 N/m×0.0548 m=118.2 NF = kx = 2156 \, \text{N/m} \times 0.0548 \, \text{m} = 118.2 \, \text{N}
Now convert this force back to mass: m=Fg=118.2 N9.8 m/s2≈12.06 kgm = \frac{F}{g} = \frac{118.2 \, \text{N}}{9.8 \, \text{m/s}^2} \approx 12.06 \, \text{kg}
✅ Final Answer: Approximately 12.06 kg of fish
🧠 Explanation (300 words)
This problem involves using Hooke’s Law, a foundational concept in physics describing how springs behave under force. When a load (like a bunch of fish) is hung from a spring, it stretches. The amount of this stretch is directly related to the force (weight) applied. Since the force depends on mass (via gravity), we can relate the extension to mass.
The first step is to determine the spring constant kk, which tells us how stiff the spring is. A stiffer spring stretches less under the same load. Using the known stretch of 5.00 cm under 11.0 kg, we find k=2156 N/mk = 2156 \, \text{N/m}. With this, we can determine how much force is required to stretch the spring to 5.48 cm.
After calculating the force for this new stretch (118.2 N), we divide by gravity to convert it to mass. The result is approximately 12.06 kg. This means the heavier load stretches the spring a bit further, in exact proportion, confirming that the spring behaves linearly as expected under Hooke’s Law.
This approach assumes the spring remains within its elastic limit, meaning it doesn’t permanently deform. If it did, the relationship would no longer be linear, and Hooke’s Law wouldn’t apply.