A force of 10N stretches an elastic spring to a total length of 20cm. If an additional force of 8N stretches the spring 4cm further, calculate the natural length of the spring.

Answer 1

Let X be the natural length of the spring.

According to Hooke's Law, the force applied to stretch a spring is directly proportional to the extension produced.

So, the force required to stretch the spring from its natural length X to 20cm is 10N.

Therefore, we can write:
10N = k(20cm - X)

Solving for k:
k = 10N / (20cm - X)

Now, the force required to stretch the spring an additional 4cm from 20cm to 24cm is 8N.

Therefore, we can write:
8N = k(24cm - X)

Substitute the value of k into the equation:
8N = (10N / (20cm - X))(24cm - X)

Simplify:
8N = (240cm - 10X) / (20cm - X)

8N(20cm - X) = 240cm - 10X
160cm - 8X = 240cm - 10X
2X = 80cm
X = 40cm

Therefore, the natural length of the spring is 40cm.