A spring 20cm long is stretched to 25cm by a load of 50N. What will be it's length when stretched by 100N assuming that the elastic limit is not reached?

Answer 1

We can use Hooke's law which states that the extension of a spring is directly proportional to the force applied, as long as the elastic limit is not reached.

Let x be the extension of the spring when a load of 50N is applied. Using Hooke's law, we have:

x/20 = 50/ƙ where ƙ is the spring constant

Solving for ƙ, we get:

ƙ = 50/(x/20) = 1000/x

Now we can use this value of ƙ to determine the extension of the spring when a load of 100N is applied:

x/20 = 100/ƙ

Substituting ƙ = 1000/x, we get:

x/20 = 100x/1000

Simplifying, we find:

x = 25cm

Therefore, the spring will be stretched to 25cm + 5cm = 30cm when a load of 100N is applied.