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1. A 455 g mass, hanging at rest on a spring, stretches the spring 22.4 cm beyond its relaxed
position. What is the spring constant of that spring?

Respuesta :

asuwafohjames

Answer:

19.9 N/m

Explanation:

From the question,

Applying Hook's law

F = Ke.................. Equation 1

Where F = Force on the spring, k = spring constant, e = extension

But the force on the spring is the weight of the mass

Therefore,

mg = ke.................. Equation 2

Where m = mass. g = acceleration due to gravity

make e the subject of the equation

e = mg/e................ Equation 3

Given: m = 455 g = 0.455 kg, e = 22.4 cm = 0.224 m,

Constant: g = 9.8 m/s²

Substitute these values into equation 3

e = (0.455×9.8)/0.224

e = 19.9 N/m

onyebuchinnaji

The spring constant of the given spring is 20 N/m.

The given parameters:

  • Mass attached, m = 455 g = 0.455 kg
  • Extension of the spring, x = 22.4 cm = 0.224 m

The spring constant is calculated by applying Hooke's law as follows;

[tex]F = kx\\\\mg = kx\\\\k = \frac{mg}{x} \\\\k = \frac{0.455 \times 9.8}{0.224} \\\\k = 20 \ N/m[/tex]

Thus, the spring constant of the spring is 20 N/m.

Learn more about Hooke's law here: https://brainly.com/question/2648431

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