A 5.0 kg mass is connected to a spring that is compressed a distance 0.10 m. A s

Question

A 5.0 kg mass is connected to a spring that is compressed a distance 0.10 m. A second, identical mass is set next to the first mass but not connected to it. When the masses are released they both are pushed toward the equilibrium position. The spring constant is k=15 N/m.
a. After passing through the equilibrium position what happens to the mass in front?
b. How far past the equilibrium point does the mass still connected to the spring travel?
c. How far is the spring compressed when this mass has completed the oscillation?

Please show detailed steps

Explanation / Answer

(a) After passing through equilibrium, the mass in front moves with constant velocity and loses contact with the attached mass.

V = 0.122 m/s

(b) As energy is halved, distance decreases by sqrt(2) times

d = 0.1/sqrt(2) = 0.071 m

(c) 0.071 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.