A spring 20cm long is stretched to 25cm by a load of 59N. What will be its length when stretched by 100N assuming that the elastic limit is not reached?

5cm/59N=X/100N solve for x.

To solve this problem, we can make use of Hooke's law, which states that the extension of a spring is directly proportional to the force applied to it. The equation for Hooke's law is given as:

F = k * ΔL

Where:
F is the force applied to the spring,
k is the spring constant,
ΔL is the change in length of the spring.

In this case, we are given the initial length of the spring (20cm), the load applied to the spring (59N), and we want to find the new length of the spring when the load is changed to 100N.

First, we need to calculate the spring constant (k). We can do this by rearranging Hooke's law equation:

k = F / ΔL

Given F = 59N and ΔL = 25cm - 20cm = 5cm, we can substitute these values into the equation:

k = 59N / 5cm

k = 11.8 N/cm

Now that we have the spring constant (k), we can find the new length of the spring when the load is 100N. Let's call this length L:

100N = k * ΔL

We need to find ΔL, the change in length of the spring. Rearranging the equation:

ΔL = 100N / k

Substituting the value of k (11.8 N/cm):

ΔL = 100N / 11.8 N/cm

ΔL = 8.47 cm

To find the new length of the spring (L), we add the change in length (ΔL) to the original length of the spring (20cm):

L = 20cm + 8.47 cm

L ≈ 28.5 cm

Therefore, the length of the spring will be approximately 28.5 cm when stretched by 100N, assuming the elastic limit is not reached.