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1.2 m long spring with a spring force of 200.0 N/m, 12.0 kg mass is attached to it, find the length
of the spring after extension after the mass is attached.
(hint: F = -6X)
(1 Point)
A) 1.232 m
B) O 1.788 m
C) 02.112 m
D) 2.788 m

Respuesta :

Answer: 1.232

Explanation:I remember having this as a hw and this was the answer correct me if I was wrong

temdan2001

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.

HOOKE'S LAW

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 [tex]L_{1}[/tex] = 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

[tex]L_{2}[/tex] = [tex]L_{1}[/tex] + e

[tex]L_{2}[/tex] = 1.2 + 0.588

[tex]L_{2}[/tex] = 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: https://brainly.com/question/12253978