How Fast Does a Stapling Gun's Ram Move at Impact?

[Cluless]

Homework Statement

A heavy-duty stapling gun uses a 0.151-kg metal rod that rams against the staple to eject it. The rod is pushed by a stiff spring called a "ram spring" (k = 35007 N/m). The mass of this spring may be ignored. Squeezing the handle of the gun first compresses the ram spring by 3.9 multiplied by 10-2m from its unstrained length and then releases it. Assuming that the ram spring is oriented vertically and is still compressed by 1.0 multiplied by 10-2m when the downward-moving ram hits the staple, find the speed of the ram at the instant of contact.

Homework Equations

U=(1/2)kx^2
(1/2)kx^2=(1/2)mv^2

The Attempt at a Solution

I have no idea what to do. My answer was 13.96 m/s but it was wrong.

 

[Delphi51]

I don't get 13.96. (1/2)kx^2=(1/2)mv^2 is not quite right because the spring still has some energy left (x = .01) when it hits.

 

[badphysicist]

Conservation of energy. U_0 + KE_0 = U_f + KE_f. Think about each of these terms and what their values are initially and finally for this situation.