Final answer:
The spring stretches by 14 meters when a 5kg mass is hung from it, as calculated using Hooke's Law. Similar principles are used to solve other problems involving spring constant calculations and deformations under weight.
Step-by-step explanation:
To determine how far the spring stretches under the weight of a 5kg mass, we use Hooke's Law, which states that the force (F) exerted by a spring is directly proportional to the extension or compression of the spring (x), and is given by the formula F = kx, where k is the spring constant. Gravity exerts a force on the mass equal to the weight of the mass (weight = mass × acceleration due to gravity), which in this case is F = 5kg × 9.8m/s2 = 49N. Thus, the spring stretch (x) can be calculated as x = F/k. Substituting the values, x = 49N / 3.5N/m = 14m.
Regarding the remaining problems, a similar process would be employed where we utilize Hooke's Law and other relevant principles of physics such as conservation of energy to solve for the force constants of springs, extension under various loads, and related measurements such as the half-kilogram marks on a scale for fish.