Question
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?
Answers
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.
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.