Question

A block of mass 0.408 kg is hung from a vertical spring and allowed to reach equilibrium at rest. As a result, the spring is stretched by 0.606 m. Find the spring constant.

Answer 1

6.598 N/m

Step-by-step explanation:

We are given that

Mass of block=0.408 kg

Displacement=x=0.606 m

We have to find the value of spring constant.

We know that

F=kx

Using the formula

`(0.408)g=k(0.606)`

g=
`9.8m/s^2`

Using the value of g

`k=(0.408* 9.8)/(0.606)=6.598`N/m

Hence, the spring constant=6.598 N/m