Question

A spring is hanging from the ceiling. When a 250 gram of mass is attached to the free end, the spring elongates by 5 cm. The spring constant (in N/m) is:

Answer 1

Answer:

k = 49 N/m

Explanation:

Given that,

Mass, m = 250 g = 0.25 kg

When the mass is attached to the end of the spring, it elongates 5 cm or 0.05 m. We need to find the spring constant. Let it is k.

The force due to mass is balanced by its weight as follows :

mg=kx

$k=\dfrac{mg}{x}\\\\k=\dfrac{0.25\times 9.8}{0.05}\\\\k=49\ N/m$

So, the spring constant of the spring is 49 N/m.