physics question- HELP?

Question

A heavy-duty stapling gun uses a 0.184-kg metal rod that rams against the staple to eject it. The rod is pushed by a stiff spring called a "ram spring" (k = 36352 N/m). The mass of this spring may be ignored. Squeezing the handle of the gun first compresses the ram spring by 3.5 10-2m from its unstrained length and then releases it. Assuming that the ram spring is oriented vertically and is still compressed by 1.4 10-2m when the downward-moving ram hits the staple, find the speed of the ram at the instant of contact.
___________ m/s

help would be appreciated, even if you can just get me started in the right direction.

Answer 1

First, calc the energy stored in the spring. It's the work done in compressing the spring.

PE = W = integral x=0 to x=d { F*dx }

Since F = kx , where x is measured from the unstrained length, this intergral is easy:

PE = W = kx^2/2 , where x is the strained length, (fix signs yourself)



It's probably easier to note that the change in PE of the spring goes into KE of the mass. So, find the difference in PE of the spring from fully compressed to the partially compressed length, equate this difference to the KE of the mass, mv^/2, and solve for v.

Answer 2

id say to use a momentum equation, an inelastic collision