# If the length of a 17 cm spring increases to 61 cm when a 4 kg weight is hanging from it, what is the spring's constant?

See Below.

#### Explanation:

We know,

The linear expansion of a elastic body is directly proportional to the force exerted on it.

So, $F \propto - \Delta x$. [The minus is to show that the body is trying to regain its original shape.]

So, $F = - k \Delta x$ [$k$ is the constant of proportionality and $k \ne 0$]

Now, Subsititute The Values in the relation.

$\textcolor{w h i t e}{\times} F = - k \Delta x$

$\Rightarrow m g = - k \left(61 - 17\right)$ [Here, $F = m g$]

$\Rightarrow 4 \times 10 = - k \times 44$ [Taking $g = 10$ $\frac{m}{s} ^ 2$]

$\Rightarrow k = - \frac{\cancel{4} \times 10}{\cancel{44}} _ 11$

$\Rightarrow k = - \frac{10}{11}$

Hence Explained.