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

Answer:

6.598 N/m

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

$k=\frac{0.408\times 9.8}{0.606}=6.598$N/m

Hence, the spring constant=6.598 N/m