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