Question

When a 4.25 kg object is placed on top of a vertical spring, the spring compresses a distance of 2.62 cm. what is the force constant of the spring?

Answer 1

The force applied to the spring is the weight of the object that compresses it, so it is equal to:
$W=mg=(4.25 kg)(9.81 m/s^2)=41.7 N$

Because of this force, the spring compresses by $x=2.62 cm=0.0262 m$. Using Hook's law,
$F=kx$,
since we know the intensity of the force (the weight W) and the compression of the spring, x, we can find k, the spring constant:
$k= \frac{F}{x}= \frac{W}{x}= \frac{41.7 N}{0.0262 m} =1591.6 N/m$