Final answer:
If we keep adding weight to a spring, eventually we reach and pass its limit of elasticity. Spring B, which has a lower force constant and therefore less stiffness, will extend more than Spring A when equal weights are suspended from both. The correct answer is A.
Step-by-step explanation:
When two springs are attached to two hooks, and each spring has a different force constant, the spring with the greater force constant will be stiffer. If equal weights are suspended from both springs, the stiffer spring (Spring A) will have less extension compared to the less stiff spring (Spring B).
According to Hooke's law, the extension of a spring is directly proportional to the force applied to it until it reaches its elastic limit, but inversely proportional to the spring's force constant (k).
Therefore, for equal weights, a spring with a higher force constant (indicating greater stiffness) will extend less than a spring with a lower force constant. This can be illustrated by envisioning a massive paperweight causing a good sag in a flexible rubber sheet; similarly, a spring with less stiffness will deform more under the same weight. The correct answer is, therefore, Spring B will have more extension than Spring A. The correct answer is A.
'Your full question was
Place the 2 single weight with a known mass on the spring( a and b) and release it. Eventually, the weight will come to rest at an equilibrium position, with the spring somewhat stretched compared to its original. If we keep adding weight to a spring b, eventually we reach and pass its limit of elasticity. Which of the following is most correct?
a. Spring B will have more extension than Spring A.
b. Spring A will have more extension than Spring B.
c. They're in equillibrium position