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
Jan 10, 2016

Answer:

#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).