last question test 1

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This question will be moved. The Forum has sections for questions for each test (GS-1 to GS-10) so this question will be moved to GS-1 Physical Sciences.

In terms of the sign convention for this problem, to be precise, we have to go step by step. Let's start from the beginning which is Newton's Second Law:

Sum of forces F = ma

Now let's take a look at the forces involved. Take the downwards direction as positive. Thus we will have a positive mg (the weight which is directed downwards). The spring, of course, is counteracting the weight by pulling upwards so we will have a - Fs. Thus:

mg - Fs = ma

but a = 0 at EQ for a spring, so we get:

- Fs = - mg

Fs = mg

now we know that Fs = -kx and that 0.5 meters is the new EQ position. The real EQ position of the spring (had there been no mass added; of course this is first year MCAT stuff so the mass of the spring is ignored) is 0.5 m up (in the negative direction) so it is -x (ie or when you plug in, put -0.5) so we can write:

Fs = -k(-x) = kx = mg

Ref:
www.ux1.eiu.edu/~cfadd/1150/15Period/Vert.html