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A 10 kilogram object suspended from the end of a vertically hanging spring stretches the spring 9.8 centimeters. At time t=0, the resulting mass-spring system is disturbed from its rest state by the force F(t)=140cos(8t). The force F(t) is expressed in Newtons and is positive in the downward direction, and time is measured in seconds.

Respuesta :

Manetho

Answer:

K= 1000 N-m

Explanation:

It is assumed that we asked to find the spring constant k of the spring

We know that under equilibrium condition

weight of the object = force applied by the spring

given m =10 Kg

x= extension in the spring = 9.8 cm

mg=kx

[tex]10\times9.8=k\times9.8\times10^(-2)[/tex]

K= 1000 N-m

fichoh

The spring constant, k of the vertically suspended spring measured in Newton meter is 1000 N/m

Given the Parameters :

  • Acceleration due to gravity, g = 9.8 m/s²

  • Mass of object, m = 10 kg

  • Extension, e = 9.8 centimeters

Using the Relation :

  • F = ke ; k = spring constant

  • Force, F = mass × acceleration due to gravity

The expression can be written thus :

mg = ke

Converting, extension to meters = (9.8/100) = 0.098 m

(10 × 9.8) = 0.098k

98 = 0.098k

k = 98/0.098

k = 1000 N/m

Therefore, the spring constant is 1000 N/m

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