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

k = 49 N/m

Step-by-step 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=(mg)/(x)\\\\k=(0.25* 9.8)/(0.05)\\\\k=49\ N/m`

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