a spring 20cm long is stretched to 25cm by a load of 50N. what will be its length when stretched by 100N assuming that the elastic limit is not reached
F=kx
F: refers to the force applied to the spring
K: the spring constant of your particular spring.
X: the distance stretched
Find the spring constant from the first statement of your question and use that to find the distance stretched by the 100N force.
1.) 10cm
To find the length of the spring when stretched by 100N, we can use Hooke's Law. Hooke's Law states that the force applied on a spring is directly proportional to the extension (change in length) of the spring.
We are given that a load of 50N stretches the spring from 20cm to 25cm. Let's calculate the constant of proportionality, which is also known as the spring constant.
The change in length of the spring is:
ΔL = 25cm - 20cm = 5cm
The force applied on the spring is:
F = 50N
Using Hooke's Law:
F = k * ΔL
where k is the spring constant.
Rearranging, we can solve for k:
k = F / ΔL
k = 50N / 5cm
k = 10 N/cm
Now, we can use the spring constant to calculate the change in length when a load of 100N is applied.
Let x be the change in length when a 100N load is applied.
100N = (10 N/cm) * x
Rearranging, we can solve for x:
x = 100N / (10 N/cm)
x = 10 cm
Therefore, when stretched by 100N, the length of the spring will increase by 10cm, and the final length will be:
Original length + change in length = 20cm + 10cm = 30cm