A spring 20cm long is stretched to 25cm by a load of 50N. What will be the length when stretched by 100N, assuming that the elastic limit is not exceeded.
100N is twice as much as 50N
so the spring will be stretched twice as far.
25-20 = 5 cm increase.
50/5 = 100/x
X = ___ cm increase.
Length = 20+x.
To find the length of the spring when stretched by 100N, we can use Hooke's Law, which states that the extension of a spring is directly proportional to the force applied to it, as long as the elastic limit is not exceeded.
First, let's calculate the spring constant, k, using the given information. The spring constant represents the stiffness of the spring and is calculated as the force divided by the extension:
k = (force applied) / (extension)
k = 50N / 0.05m
k = 1000 N/m
Now we can use this spring constant to calculate the extension of the spring when a force of 100N is applied:
extension = (force applied) / (spring constant)
extension = 100N / 1000 N/m
extension = 0.1m
Finally, to find the length of the spring when stretched by 100N, we add the extension to the original length of the spring:
length = original length + extension
length = 0.2m + 0.1m
length = 0.3m
Therefore, the spring will be 30cm long when stretched by 100N, assuming the elastic limit is not exceeded.