A 10 kilogram object suspended from the end of a vertically hanging spring stretches the spring 9.8 centimeters. At time t

Full Question

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) = 20 cos (8t). The force F (t) is expressed in Newtons and is positive in the downward direction, and time is measured in seconds.

Determine the spring constant k.

Answer:

Spring Constant = 1000Nm^-1

Step-by-step explanation:

Given

Mass = 10kg

Spring Extension = 9.8cm

Spring Constant K = Force/Extension

Force = Mass * Acceleration

Force = 10 * 9.8 = 9.8N

Spring Extension = 9.8 cm = 9.8 / 100 = 0.098 m

Spring constant = 98 / 0.098

Spring Constant = 1000Nm^-1