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?

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

inverse v=i*r

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