A spring is compressed 27.8 cm when a force of 76.5 N is applied. How fast will an 5.9 kg mass move if the spring is compressed 49.8 cm and released with the mass attached?
The correct answer and explanation is:
To solve this problem, we use Hooke’s Law and the Conservation of Energy principle.
Step 1: Find the spring constant kk
Hooke’s Law relates force and displacement for a spring: F=kxF = kx
Where:
- F=76.5 NF = 76.5 \, \text{N}
- x=27.8 cm=0.278 mx = 27.8 \, \text{cm} = 0.278 \, \text{m}
Rearrange to solve for kk: k=Fx=76.50.278≈275.27 N/mk = \frac{F}{x} = \frac{76.5}{0.278} \approx 275.27 \, \text{N/m}
Step 2: Use energy conservation to find speed
When the spring is compressed and then released, all the potential energy stored in the spring is converted into the kinetic energy of the attached mass: 12kx2=12mv2\frac{1}{2}kx^2 = \frac{1}{2}mv^2
Where:
- k=275.27 N/mk = 275.27 \, \text{N/m}
- x=49.8 cm=0.498 mx = 49.8 \, \text{cm} = 0.498 \, \text{m}
- m=5.9 kgm = 5.9 \, \text{kg}
- v=final speedv = \text{final speed}
12(275.27)(0.498)2=12(5.9)v2\frac{1}{2}(275.27)(0.498)^2 = \frac{1}{2}(5.9)v^2
Simplify: 12(275.27)(0.248004)=12(5.9)v2\frac{1}{2}(275.27)(0.248004) = \frac{1}{2}(5.9)v^2 34.126≈2.95v234.126 \approx 2.95v^2 v2=34.1262.95≈11.57v^2 = \frac{34.126}{2.95} \approx 11.57 v≈11.57≈3.40 m/sv \approx \sqrt{11.57} \approx 3.40 \, \text{m/s}
✅ Final Answer:
3.40 m/s\boxed{3.40 \, \text{m/s}}
📘 Explanation (300 words):
In this problem, we analyze how energy stored in a compressed spring is converted to motion. When a spring is compressed, it stores elastic potential energy. According to Hooke’s Law, the force needed to compress a spring is directly proportional to the displacement: F=kxF = kx, where kk is the spring constant. Using the initial data — a force of 76.5 N causing a compression of 0.278 m — we calculate the spring constant as approximately 275.27 N/m.
When the spring is compressed further (to 49.8 cm) and released with a 5.9 kg mass attached, that potential energy becomes kinetic energy. The formula for potential energy stored in a spring is: PE=12kx2PE = \frac{1}{2}kx^2
This energy is transformed into kinetic energy of the moving mass: KE=12mv2KE = \frac{1}{2}mv^2
Equating both (since energy is conserved), we find the final speed of the mass. Plugging in values and solving gives a final velocity of 3.40 m/s.
This principle is central in physics and engineering, particularly in systems involving springs or elastic materials, from car suspensions to toys.