Question

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?

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.