Question involving spring constant

AI Thread Summary
The discussion revolves around calculating the new length of a spring when a 65 kg weight is applied, using a spring constant of 12000 N/m. The user correctly calculates the force exerted by the weight as 637.65 N and applies Hooke's Law to find the change in length of the spring. The calculated extension of the spring is approximately 0.053 m, which initially seems small. However, further reflection reveals that this extension is reasonable when converted to centimeters, as it represents a minor adjustment relative to the spring's original length. The conclusion emphasizes the importance of perspective in interpreting the results of spring mechanics.
Phenom66
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Homework Statement


We have a spring that is 30cm with a spring constant of 12000N/m
and there is a 65 kg weight put onto this spring. What is the new length of the spring?


Homework Equations


f = k *changeof* x


The Attempt at a Solution


I tried to take 65kg times 9.81 to find Newtons (which is 637.65N)
so therefore 637.65 = 12000 *change of* x
and x = 637.65/12000
But something feels fishy about this, a spring doesn't move down just .053 m
 
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Phenom66 said:

Homework Statement


We have a spring that is 30cm with a spring constant of 12000N/m
and there is a 65 kg weight put onto this spring. What is the new length of the spring?


Homework Equations


f = k *changeof* x


The Attempt at a Solution


I tried to take 65kg times 9.81 to find Newtons (which is 637.65N)
so therefore 637.65 = 12000 *change of* x
and x = 637.65/12000
But something feels fishy about this, a spring doesn't move down just .053 m

Think about it for a moment longer. How big is 637.65 N compared to 12000 N?

Then convert .053 m into centimeters. Now does it seem so unreasonable?
 

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