Homework Help: phsyics

Posted by Anonymous on Thursday, November 5, 2009 at 11:43pm.

A mass, m1 = 9.36 kg, is in equilibrium while connected to a light spring of constant k = 112 N/m that is fastened to a wall.

A second mass, m2 = 6.71 kg, is slowly pushed up against mass m1, compressing the spring by the amount A = 0.201 m

The system is then released, and both masses start moving to the right on the frictionless surface. When m1 reaches the equilibrium point, m2 loses contact with m1 and moves to the right with speed v. Determine the value of v.

phsyics - drwls, Friday, November 6, 2009 at 5:58am
At the spring's equilbrium point, all of the stored potential energy of the compressed spring becomes kinetic energy of both masses. No further force is applied by the spring at that instant, and then the masses separate as the spring force reverses direction.

The V (of both masses) at that instant is given by

(1/2) k A^2 = (1/2)(m1 + m2) V^2

Solve for V

Answer this Question

Related Questions

physics - A mass, m1 = 5.5 kg, is in equilibrium while connected to a light ...
physics - A light spring of constant k = 170 N/m rests vertically on the bottom ...
physics help please... - A light spring of constant k = 170 N/m rests vertically...
Physics - A light spring of constant k = 158 N/m rests vertically on the bottom ...
physics - A light spring of spring constant k = 160 N/m rests vertically on the ...
Physics Problem - A 3.00-kg mass is fastened to a light spring that passes over ...
Physics - A spring, (spring constant of 400 N/m) is displaced .75 m from its ...
Physics - A block of mass m1 = 22.0 kg is connected to a block of mass m2 = 40.0...
physics - A light spring of constant k = 156 N/m rests vertically on the bottom...
physics - Janet wants to find the spring constant of a given spring, so she ...

For Further Reading

First Name:
School Subject:
Answer: