A heavy-duty stapling gun uses a 0.146 kg metal rod that rams against the staple

Question

A heavy-duty stapling gun uses a 0.146 kg metal rod that rams against the staple to eject it. The rod is attached and pushed by a stiff spring called a "ram spring" (k = 31325 N/m). The mass of this spring may be ignored. Squeezing the handle of the gun first compresses. The ram spring is compressed by 3.00 10-2 m from its unstrained length and then releases from rest. Assuming that the ram spring is oriented vertically and is still compressed by 4.80 10-3 m when the downward-moving ram hits the staple, find the speed of the ram at the instant of contact.

Explanation / Answer

U = potential energy of spring = 0.5*k*x2

Let velocity of ram is v m/sec.
The amount of energy transferred is proportional to the change in displacement of the spring.
Spring displacement changed by (3.0x10-2 - 4.80x10-3) m = 2.52x10-2

Energy transferred to ram = 0.5x31325x(2.52x10-2)2

                                     E = 9.9463 J

E = 0.5*m*v2

v2 = 9.9463/(0.5*0.146)

   v2 = 136.250

v = 11.67265 m/s

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