Hooke's Law: Finding the Spring Constant with a 50g Mass and 7.0cm Stretch

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To find the spring constant using Hooke's Law, the total mass affecting the spring is 70g, which includes the initial 50g and the additional 20g. The formula mg = kx applies, where m is the total mass, g is the acceleration due to gravity, k is the spring constant, and x is the stretch of the spring. The confusion arises from whether to consider the total mass or just the additional mass causing the stretch. The key indicates that the additional mass is 20g, which refers to the increase in stretch of 7.0cm, not the total mass. Correctly applying these values leads to the calculation of the spring constant.
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Homework Statement




A 50g mass hangs at the end of a hookean spring. When a 20g more is added to the spring, it stretches 7.0cm more. Find the spring constant.

The Attempt at a Solution



Isn't it just

mg = kx

mg/x = k

Shouldn't m = 70g instead of 20g?

My key says m = 20g?
 
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flyingpig said:
Shouldn't m = 70g instead of 20g?
That would give the total elongation from the unweighted position, not the additional elongation.
 

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