Calculating the Total Length of a Spring with Varying Masses

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To calculate the total length of a spring with varying masses, the initial setup involves a 6.00 kg mass that extends the spring to 0.250 m, with a spring constant of 975.0 N/m. Using Hooke's Law, the extension caused by the 6.00 kg mass is found to be 0.0603 m. When a 15.20 kg mass is applied, the extension is calculated as 0.153 m. To find the total length of the spring with the 15.20 kg mass, the unstretched length of the spring is determined by subtracting the extension from the total length. The final calculation confirms the correct total length of the spring when the larger mass is suspended.
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A 6.00 kg mass suspended from a spring with spring constant, k = 975.0 N/m, extends it to a total length of 0.250 m. Find the total length of the spring when a 15.20 kg mass is suspended from it.
I drew the free body diagram to get,
ma= kx- mg.. since a= 0, kx=mg
solving for x gave me mg/k
So I did (6.00)(9.8)/ 975.0 = .0603 m
Then I did (15.2)(9.8)/ 975.0 = .153 m
I didn't really know where to go from here.. do I subtract the values? Or am I even going about this in the right way?
Thanks
 
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Punchlinegirl said:
A 6.00 kg mass suspended from a spring with spring constant, k = 975.0 N/m, extends it to a total length of 0.250 m.
Hint: Find the unstretched length of the spring.
 
so to find the unstretched length I use Hooke's Law
F= -kd
So it would be like what I had before, mg/k= d
(6.00)(9.8)/ 975 = .0603
since the total length is .250, I do .250-.0603 to get .1990 m.
Is this right?
 
Right idea, but check the arithmetic.
 
Ok I figured it out. Thanks Doc Al!
 
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