An ornament of mass 40.0 g is attached to a vertical ideal spring with a force constant (spring constant)of 20.0 N/m.The ornament is then lowered very slowly until the spring stops stretching.How much does the spring stretch?
A)0.00200 m
B)0.0196 m
C)0.0816 m
D)0.800 m
E)0.200 m

Question

Quizplus · Accepted Answer

We can use Hooke's Law, which states that the force exerted by the spring is equal to the spring constant times the displacement of the spring from its equilibrium position: F = -kx (- sign indicates that the force is in the opposite direction of the displacement). At equilibrium, the force exerted by gravity on the mass is balanced by the force exerted by the spring, so we can write: Fg = -kx. Solving for x, we get: x = -Fg/k = -(mg)/k = - (0.040 kg)(9.81 m/s^2)/(20.0 N/m) = -0.0196 m. The negative sign is just indicating that the displacement is in the opposite direction of the gravitational force, so we take the absolute value to get: x = 0.0196 m.

An Ornament of Mass 40