if spring 20cm is stretch to 25cm by a load of son, what will be its length when stretched by 100N assuming that the elastic limit is not reach

stretches 5 cm by whatever son is

(100/son) * 5 cm + 20 cm

500son

To answer this question, we need to understand the concept of Hooke's Law, which relates the force applied to a spring to the extension it experiences. Hooke's Law states that the force applied to a spring is directly proportional to the extension it undergoes, as long as the elastic limit is not reached.

Here's how we can use Hooke's Law to solve the problem:

1. We are given that a spring initially has a length of 20cm (or 0.2m) and is stretched to 25cm (or 0.25m) by a load. The extension of the spring can be calculated as follows:

Extension = Final Length - Initial Length
= 0.25m - 0.2m
= 0.05m

2. Now, we can use Hooke's Law to find the spring constant, which relates the force applied to the extension:

Force = Spring Constant × Extension

Rearranging the equation, we have:

Spring Constant = Force / Extension

3. We know that the force applied to stretch the spring to 25cm is unknown, so we cannot directly calculate the spring constant. However, we are given that the extension is 0.05m, and the force applied is 100N. We can solve for the spring constant as follows:

Spring Constant = 100N / 0.05m
= 2000 N/m

4. Finally, to find the length of the spring when stretched by 100N, we can rearrange Hooke's Law and solve for the unknown length:

Length = Initial Length + Extension

Length = 0.2m + (Force / Spring Constant)

Length = 0.2m + (100N / 2000 N/m)
= 0.2m + 0.05m
= 0.25m

Therefore, when the spring is stretched by 100N, its length will be 25cm (or 0.25m).