A 10-kg mass, hung by an ideal spring, causes the spring to stretch 2.0 cm. what is the spring constant (force constant) for this spring?

Question

Physics

Jordin Carter

8 December, 12:07

A 10-kg mass, hung by an ideal spring, causes the spring to stretch 2.0 cm. what is the spring constant (force constant) for this spring?

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Answers (2)

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Owens · Answer 1

From Hooke's law the spring constant k is given by F = ke; then k = F/e. But F = Ma = Mg = 10 * 9.8 = 98N where g = 9.8ms^ (-2) Extension e = 2cm = 0.02m. So k = 98/0.02 = 4900 N/m.

Stephen Buck · Answer 2

The computation for this problem is calculated by:F = kx

F = (10) (9.8)

x =.02 meters after converting from cm to m

98 = (k) (.02) So, k=4900 N m.

Then back to cm by dividing by 100 which is now 49 N cm would be the spring constant for this problem.