Question

A steel spring is hung vertically. Using a ruler, the lower end of the spring was measured to be 25.0cm from its upper end. However, when an object of mass 0.500kg it attached to the lower end of the spring the spring stretches to a new length of 45.0cm.

With the aid of a diagram and with a detailed explanation of your reasoning determine:

(a) The stiffness constant k of the spring.

(b) The additional weight needed to extend the spring to 50.0cm.

(c) The force needed to extend the spring by 50.0mm from its natural length.

Answer 1

M*g = 0.5 * 9.8 = 4.9 N. = Wt. of object.

a. k = F/d = 4.9N./(0.45-0.25)m = 24.5 N/m.

b. 24.5N./m * (0.50-0.25)m = 6.125 N.
6.125 - 4.9 = 1.23 N. addition wt.

Answer 2

c. F = 24.5N/m * 0.05m = 1.23 N.

Answer 3

Henry, the problem I am trying to solve here is the explanation part. How am I meant to provide such a detailed explanation when the majority of the methods make up the decription of the answer?