Physics
posted by Leah on .
A heavyduty stapling gun uses a 0.162kg metal rod that rams against the staple to eject it. The rod is pushed by a stiff spring called a "ram spring" (k = 32657 N/m). The mass of this spring may be ignored. Squeezing the handle of the gun first compresses the ram spring by 3.0 102m from its unstrained length and then releases it. Assuming that the ram spring is oriented vertically and is still compressed by 1.3 102m when the downwardmoving ram hits the staple, find the speed of the ram at the instant of contact.

use this equation
v = x square root(k/m)
and the first x
v=?
x=3.0 x 102 m
k=32657 N/m
m=0.162 kg
v = x square root(k/m)
= (0.030m) x ((square root of)32657/0.162
=(0.030m)(square root of 201586.4198)
=(0.030m)(448.9837634)
=13.4695129 m/s
= 13 m/s