A 31.0 cm long spring is hung vertically from a ceiling and stretches to 36.9 cm when an 8.00 kg mass is hung from its free end. (a) Find the spring constant (in N/m).
 

(b) Find the length of the spring (in cm) if the 8.00 kg weight is replaced with a 185 N weight.

The length of the spring is 44.92 cm

Explanation:

Hooke's Law

Suppose a spring of constant k and natural length x0. If a force F is applied to the spring and it stretches to a distance x1. Hooke's Law states that:

Where x is the elongation of the spring:

We are given the characteristics of a spring of x0=31 cm. When a mass of m=8 Kg is hung from the spring, it stretches to x1=36.9 cm. We need to calculate the force of the mass of 8 Kg. It can be done by calculating the weight:

The elongation of the spring is

Converting to meters:

(a)

From Hooke's Law, we solve for k:

(b) With the value of k, the equation for the spring is:

Now if a weight of F=185 N is hung from the spring, the elongation is:

Thus, the length of the spring is:

The length of the spring is 44.92 cm

not too sure thinking it will always be directed towards the centre of the earth. could be wrong not sure

increase masses, decrease separation