This scenario makes use of Hooke’s Law, which states in equation form: F=-kx, where F is the restoring force, -k is the spring constant, and x is the displacement from equilibrium.
Let’s list our given information and then plug it into the equation to solve for -k: the spring constant.
• We are given that a force of 30N stretches the spring, and therefore, the restoring force must be -30N to cancel out the pulling force, resulting in zero net force since 30N-30N=0N.
• We also know the spring is displaced .73m from the equilibrium (starting point).
• We now have F (the restoring force) and x (displacement from equilibrium). F=-30N and x=.73m. Now let’s plug this information into Hooke’s Equation and solve for -k: the spring constant (the rate at which force must be applied over a distance to restore the spring back to equilibrium).
Hooke’s Equation:
F=-kx
Substituting in for variables:
(30N)=-k(.73m)
Divide both sides by .73m:
30N/.73m=-k
Divide:
41.095=-k
Apply the symmetric property:
-k=41.095N/m
Therefore, the answer is choice A.) 41N/m