If the length of a 21 cm spring increases to 57 cm when a 5 kg weight is hanging from it, what is the spring's constant?

1 Answer
Truong-Son N.
Dec 26, 2015

We should expect k to be somewhere in the hundreds for typical springs.

This is simply looking at the equation

\mathbf(F = -kDeltay)

...for when an object is hanging off a spring.

So, the only force acting on the spring is the force \mathbf(F_g) due to gravity \mathbf(g). The generic force F, then, is F_g. Thus, with g < 0 (where down is negative):

F_g = -kDeltay

mg = -k(y_f - y_i)

(mg)/(y_i - y_f) = k

color(blue)(k) = (("5 kg")(-"9.80665" cancel"m""/s"^2))/("0.21" cancel"m" - "0.57" cancel"m")

= color(blue)("136.20 kg/s"^2) or color(blue)("N/m")

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