Question

A 250 g object is hung onto a spring. It stretches 18 cm. Find the spring's spring constant

Answer 1

Answer: 13.61 N/m

Hooke's law establishes that the elongation of a spring is directly proportional to the modulus of the force $F$ applied to it, as long as the spring is not permanently deformed:

$F=k (x-x_{o})$    (1)

Where:

$k$ is the elastic constant of the spring. The higher its value, the more work it will cost to stretch the spring.

$x_{o}$ is the length of the spring without applying force.

$x$ is the length of the spring with the force applied.

According to this, we have a spring where only the force due gravity is applied.

In other words, the force applied is the weigth $W$ of the block:

$W=m.g$   (2)

Where $m=250g=0.25kg$ is the mass of the block and $g=9.8\frac{m}{s^{2}}$  is the gravity acceleration.

$W=(0.25kg)(9.8\frac{m}{s^{2}})$   (3)

$W=2.45N$   (4)

Knowing the force applied $W$ and $x=18cm=0.18m$ and $x_{o}=0$, we can substitute the values in equation (1) and find $k$:

$W=k (x-x_{o})$    (5)

$2.45N=k (0.18m-0m)$    (6)

Finally:

$k=13.61\frac{N}{m}$