If the length of a 42 cm spring increases to 87 cm when a 1 kg weight is hanging from it, what is the spring's constant?

1 Answer
Trevor Ryan.
Jan 10, 2016

k=21,778N//m

Explanation:

Hooke's Law states that the elongation or compression of an elastic spring is directly proportional to the resultant force acting on it and in the direction of the resultant force.
vecF = k vecx.

So in this case, the resultant force is the weight force W=mg=1xx9,8=9,8N.

The elongation in this case is x=87-42=45cm, which is 0,45m in SI units.

Therefore by Hooke's Law, the spring stiffness constant may be given by

k=F/x=(9,8N)/(0,45m)=21,778N//m.

(This is quite a small value, indicating a very elastic spring, easy to stretch and compress. Not very stiff).

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