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From Hooke's law, the length of the spring after extension and after the mass is attached is 1.788 meters. Option B is the answer.
<h3>HOOKE'S LAW</h3>Hooke's law state that in an elastic material, the force applied is directly proportional to the extension provided that the elastic limit is not exceeded.
Given that a 12.0 kg mass is attached to 1.2 m long spring with a spring constant of 200.0 N/m.
The given parameters are:
- mass m = 12kg
- Initial length
= 1.2 m
- Spring constant K = 200 N/m
- Extension e = ?
According to Hooke's law
F = Ke
But F = mg
mg = Ke
Substitute all the parameters into the formula to get extension e
12 x 9.8 = 200e
e = 117.6 / 200
e = 0.588 m
The length of the spring after extension and after the mass is attached will be
=
+ e
= 1.2 + 0.588
= 1.788 m
Therefore, the correct answer is option B because the length of the spring after extension and after the mass is attached is 1.788 meters.
Learn more about Hooke's law here: brainly.com/question/12253978