A mass, m1 = 5.5 kilogram, is in equilibrium while connected to a light spring o

Question

A mass, m1 = 5.5 kilogram, is in equilibrium while connected to a light spring of constant k = 140 N/m that is fastened to a wall (see a). A second mass, m_2 = 8.6 kg, is slowly pushed up against mass m_1, compressing the spring by the amount A_i = 0.1 m (see b). The sysytem is then released, causing both masses to start moving to the right on the frictionless surface. When m_1 is at the equilibrium point, m_2 loses contact with m_1(see c) and moves to the right with speed v_max. Determine the vaule of v_max. Answer in units of m/s.

Explanation / Answer

by conservation of energy

0.5*k*X2 = 0.5*(m1+m2)*V2max

Vmax=X*(k/(m1+m2))0.5 = 0.315 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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