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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.

Respuesta :

lublana

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

[tex](0.408)g=k(0.606)[/tex]

g=[tex]9.8m/s^2[/tex]

Using the value of g

[tex]k=\frac{0.408\times 9.8}{0.606}=6.598[/tex]N/m

Hence, the spring constant=6.598 N/m

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