A heavy-duty stapling gun uses a 0.162 kg metal rod that rams against the staple

Question

A heavy-duty stapling gun uses a 0.162 kg metal rod that rams against the staple to eject it. The rod is pushed by a stiff spring called a "ram spring" (k = 36427 N/m). The mass of this spring may be ignored. Squeezing the handle of the gun first compresses the ram spring by 3.20 10-2 m from its unstrained length and then releases it. Assuming that the ram spring is oriented vertically and is still compressed by 8.90 10-3 m when the downward-moving ram hits the staple, find the speed of the ram at the instant of contact.

Explanation / Answer

Change in spring energy = rise in potential energy of mass + rise in KE of mass So, 14.958 J = 0.0428 J + 0.5 * 0.164 * v^2 So, V = 13.486 m/s

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