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**1. Homework Statement**

"A 1.50 kg object hangs motionless from a spring with a force constant of k = 250 N/m. How far is the spring stretched from its equilibrium length?"

**2. Homework Equations**

F = -kx

W = mg

The teacher said to call the force the spring pulls up with F

_{T}.

**3. The Attempt at a Solution**

Hello, I was wondering if somebody could help me understand my errors. I started the word problem by creating this diagram:

I then proceeded to sum the forces in the y-axis:

∑F

_{y}: F

_{T}+ W = ma

and substituted to:

-kx + mg = ma

a = 0 so I re-arranged the equation to say

∑F

_{y}: -kx = -mg

I plugged the numbers from the problem into the equation as follows:

-(250)(x) = -((1.5)(-9.8))

I solved for x and arrived at the solution of

x = -0.0588 m

My teacher told me that my solution was correct, but my work was wrong. He said the work should look like this:

∑F

_{y}: F

_{T}- W = ma

-kx - mg = ma

a = 0

∑F

_{y}: -kx = mg

-(250)(x) = (1.5)(9.8)

x = -0.0588 m

He explained that because the weight is in the negative (downward) direction I should subtract mg from -kx, and that I should plug in 9.8 for g instead of -9.8. He says that my way of adding together F

_{T}and W while plugging in -9.8 for g is incorrect because it "does not show the direction". I am confused about why I can not add together the two forces and plug in -9.8 for g, I must be missing something. He also explained that "The direction is taken care of in the free body diagram" what exactly does this mean? If direction is already taken care of in the FBD then how come we are even subtracting weight in the first place? I was wondering what exactly the free body diagram does for me and also why I am wrong in making g negative. Would the negative direction of the acceleration (g) not mean that weight is also negative? I guess what I am trying to say is, what makes my incorrect method different from the correct method, and why would I not plug in a = -9.8? Thank you for your help.

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