# Spring compression distance

Given: Load of sacks have total weight of 2250 N and each sack weighs 225 N, how far will each of 2 springs in one system be compressed (spring coeff = 2200 N/m) when a sack is put on it?

Energy init +work=final Energy

2250=1/2(4400)x^2
1.02=x^2
x=1.01

Not sure where to go from here.

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turin
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Given: Load of sacks have total weight of 2250 N and each sack weighs 225 N, how far will each of 2 springs in one system be compressed (spring coeff = 2200 N/m) when a sack is put on it?

Energy init +work=final Energy
I wouldn't use energy. There is another (even simpler) spring relation that I would use. BTW, I don't know what the problem means by "one system".

Then using Us=1/2kx^2
I have 1/2(2200)x^2=2250
x=1.43

One system meaning that there are 2 springs working identically to each other.

turin
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Is there a reason why you can't use Hooke's law?

One system meaning that there are 2 springs working identically to each other.
I still don't know what that means.

Using Hooke's law Fs=-kx
2250N=(-2200N/m)x
x=1.02m
1.02m/2 springs =0.5 m

turin
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2250N=(-2200N/m)x
x=1.02m
What is the meaning of this x value? Can you interpret it physically?

The x value is the total amount that the springs would be compressed by a force of 2250 N.

turin
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The x value is the total amount that the springs would be compressed by a force of 2250 N.
And is that the condition that the problem is asking for?

I didn't post the picture of the springs. It shows them as being parallel to each other with a platform on top. Sorry. My bad. I should be using them together. The link helped as well.

Now, using Fspring=-kx
2250=(-)(2200+2200)x
x=0.5 m

turin
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Now, using Fspring=-kx
2250=(-)(2200+2200)x
x=0.5 m
I ask again, is that the x that the problem is asking for? Hint: it has the same interpretation as the last time I asked.

Yes, they are asking for x.

turin
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Yes, they are asking for x.
Nope.