Question

A heavy-duty stapling gun uses a 0.147-kg metal rod that rams against the staple to eject it. The rod is pushed by a stiff spring called a "ram spring" (k = 32800 N/m). The mass of this spring may be ignored. Squeezing the handle of the gun first compresses the ram spring by 4.44 cm from its unstrained length and then releases it. Assuming that the ram spring is oriented vertically and is still compressed by 1.53 cm when the downward-moving ram hits the staple, find the speed of the ram at the instant of contact.

Answer 1

Answer:

 v = 13.75 m/s

Explanation:

mass of rod (m) = 0.147 kg

spring constant (k) = 32800 N/m

initial compression (Y1) = 4.44 cm = 0.0444 m

final compression (Y2) = 1.53 cm = 0.0153 m

find the speed at the instant of contact

kinetic energy of a spring = elastic potential energy

0.5m$v^{2}$ = 0.5k$x^{2}$

$v^{2}$ = $\frac{0.5k[tex]x^{2}$}{0.5m} \[/tex]

where x = Y1 - Y2

x = 0.0444 - 0.0153 = 0.0291 m

$v^{2}$ = $\frac{0.5 x 32800 x [tex]0.0291^{2}$}{0.5 x 0.147} \[/tex]

        $v^{2}$ = 188.95

       v = 13.75 m/s