An ideal spring obeys Hooke'* law, F = -kx. A mass of 0.50 kilogram hung vertically from this spring stretches the spring 0.075 meter. The value of the force constant for the spring is most nearly
(A) 0.33 N/m
(B) 0.66 N/m
(C) 6.6 N/m
(D) 33 N/m
(E) 66 N/m
Correct option (E) 66 N/m
Force is provided by the weight of the mass (mg). Simply plug into F = k∆x, mg = k∆x and solve.
