Final answer:
Using Hooke's Law, the spring stretches to 20 meters when a 10-kilogram weight is attached, derived from the relationship between the force applied by the weight and the spring constant.
Step-by-step explanation:
The subject of this question is Physics, specifically related to Hooke's Law and the extension of springs under an applied load. The problem describes a spring stretching 8 meters when a 4-kilogram weight is attached. We are asked to determine the length the spring stretches when a 10-kilogram weight is attached.
To solve this, we use Hooke's Law, which states that the extension (x) of a spring is directly proportional to the load (F) applied to it, assuming the elasticity limit is not exceeded. From the problem, we know that the spring stretches 8 meters when a 4-kilogram weight is attached. The force exerted by the weight due to gravity (F) is its mass multiplied by the acceleration due to gravity (g): F = mg, where m is mass and g is approximately 9.8 m/s2.
For the 4-kilogram weight: F1 = 4 kg × 9.8 m/s2 = 39.2 N.
Extension (x) = 8 meters.
Therefore, the spring constant (k) can be calculated as k = F1 / x = 39.2 N / 8 m = 4.9 N/m.
Now, with a 10-kilogram weight, the force would be F2 = 10 kg × 9.8 m/s2 = 98 N.
Using the same spring constant (k), the new extension (x2) can be found:
x2 = F2 / k = 98 N / 4.9 N/m = 20 meters.
The spring will stretch to 20 meters when a 10-kilogram weight is attached, thus the correct answer is option c. 20 meters.