If the length of a 24 cm spring increases to 47 cm when a 5 kg weight is hanging from it, what is the spring's constant?

1 Answer
David G.
Jan 18, 2016

Better to rewrite the question in metres: "0.24 m spring increases 0.47 m..." Spring constant, k = F/(Delta d) = (m*g)/(Delta d) = (5*9.8)/(0.47-0.24) = 21.3 Nm^-1

Explanation:

The spring constant, k, of a spring is expressed in newton per metre (Nm^-1): for each additional N of force the spring expands by k (m).

It's important to use correct SI units for the quantities: centimetre is not a base SI unit, metre is, so convert the cm to m.

I used Delta d to represent the change in length of the spring - final length of 0.47 m minus initial length 0.24 m.

The force acting on the spring is the weight force of the mass, which is its mass, 5 kg times g = 9.8 Nkg^-1 (more often written as ms^-2, but this is an exactly equivalent unit that makes more sense in this context).

Calculating all this leads to a spring constant equal to 21.3 Nm^-1.

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