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A spring with a 10-kg mass and a damping constant 15 can be held stretched 2 meters beyond its natural length by a force of 6 newtons. Suppose the spring is stretched 4 meters beyond its natural length and then released with zero velocity. Suppose the spring is stretched 4 meters beyond its natural length and then released with zero velocity.

Required:
Find the position of the mass at any time t.

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Answer:

Explanation:

Given that:

mass = 10kg

damping constant C = 15 kg/s

length = 2 m

Force F = 6N

Using the Hooke's law:

F = kx

6 = 15x

k = 6 N /2 m

spring constant k = 3 N/m

For the critical damping

C² - 4k*m= 0

m = C²/4k

m = (15)²/4(3) kg

m = 225/12 kg

m = 18.75 kg

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