Final answer:
The resulting displacement of the mass-spring system when a 3.2-pound mass is attached is 1.6 feet, calculated by dividing the force due to gravity by the spring constant. The damping force affects the motion, not the static displacement.
Step-by-step explanation:
The question asks for the resulting displacement of a mass-spring system when a mass is attached, and the system is placed in a medium offering damping force. Given that a force of 2 pounds stretches the spring 1 foot, we can determine the spring constant to be k = 2 lbs/ft. Attaching a 3.2-pound mass to the spring results in a force due to gravity of F = mg, where g is the acceleration due to gravity in pounds-force per pound-mass, taken as 32.17405 ft/s². Since 1 pound-mass exerts a force of 1 pound-force under gravity, the downward force F = 3.2 lbs.
The static displacement, x, is given by F = kx. Substituting the given values, we have 3.2 lbs = (2 lbs/ft) × x, resulting in x = 1.6 feet. However, the system offers a damping force of 0.8, which implies a counteracting force to the motion. While the question mentions a damping force, it doesn't specify how it affects the system statically. Typically, damping force affects the motion of the system, not the static displacement. Therefore, the damping would not change the static displacement caused by the attached mass, and the displacement of 1.6 feet is correct before considering any dynamic factors.