A heavy-duty stapling gun uses a 0.144-kg metal rod that rams against the staple

Question

A heavy-duty stapling gun uses a 0.144-kg metal rod that rams against the staple to eject it. The rod is pushed by a stiff spring called a "ram spring" (k = 38900 N/m). The mass of this spring may be ignored. Squeezing the handle of the gun first compresses the ram spring by 4.76 cm from its unstrained length and then releases it. Assuming that the ram spring is oriented vertically and is still compressed by 1.04 cm when the downward-moving ram hits the staple, find the speed of the ram at the instant of contact. Since the spring is vertical, we have to account for gravitational potential energy as well as kinetic and elastic potential energy. Thus, the energy can be written: E=(1/2)mv^2+mgh+(1/2)kx^2. Set Ei = Ef and solve for vf (v final). Round to three significant figures.

Explanation / Answer

A heavy-duty stapling gun uses a 0.175-kg metal rod that rams against the staple to eject it. The rod is pushed by a stiff spring called a "ram spring" (k = 34421 N/m). The mass of this spring may be ignored. Squeezing the handle of the gun first compresses the ram spring by 3.7 10-2m from its unstrained length and then releases it. Assuming that the ram spring is oriented vertically and is still compressed by 1.2 10-2m when the downward-moving ram hits the staple, find the speed of the ram at the instant of contact. U = 0.5*k*x^2 U = potential energy of spring The amount of energy transferred is proportional to the change in displacement of the spring. Spring displacement changed by (3.7 - 1.2) = 2.5*10^-2 Energy transferred to ram = 0.5*34421*(2.5*10^-2)^2 = 10.7566 J Ek = 0.5*m*v^2 v^2 = 10.7566/(0.5*0.175)

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