Question

A heavy-duty stapling gun uses a 0.179 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 = 37107 N/m). The mass of this spring may be ignored. Squeezing the handle of the gun first compresses. The ram spring is compressed by 3.20 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 1.35 10-2 m when the downward-moving ram hits the staple, find the speed of the ram at the instant of contact.

Answer 1

v = 13.22 m/s

Step-by-step explanation:

mass of rod (M) = 0.179 kg

spring constant (k) = 37107 N/m

initial compression (Ei) = 3.2 x 10^{-2} m

final compression (Ef) = 1.35 x 10^{-2} m

acceleration due to gravity (g) = 9.8 m/s^{2}

from the conservation of energy, the total energy in the system before impact = the total energy in the system after impact

`0.5.mv^(2) + mg(hf) + 0.5k(Ef)^(2) = 0.5.mu^(2) + mg(hi) + 0.5k(Ei)^(2)`

where

• m = mass = 0.179 kg
• v = final speed at the instant of contact
• u = initial speed = 0 since it was initially at rest
• hf = final height
• Hi = initial height
• Ef = final compression = 1.35 x 10^{-2} m

• Ei = initial compression = 3.2 x 10^{-2} m
• k = spring constant = 37107 N/m
• g = acceleration due to gravity = 9.8 m/s^{2}
• since u = 0, the equation now becomes

`0.5.mv^(2) + mg(hf) + 0.5k(Ef)^(2) = mg(hi) + 0.5k(Ei)^(2)`

now we rearrange the equation above to make v the subject of the formula

`v= \sqrt{(k(Ei^(2)-Ef^(2)))/(m) + 2g(hi - hf) }`

• hi - hf = change in height = change in extension =
  `(3.2x10^(-2)) -(1.35x10^(-2)) = 0.0185 m`
• now substituting all required values into the equation above we have

`v= \sqrt{(37107((3.2x10^(-2))^(2) -(1.35x10^(-2))^(2) ))/(0.179)+2x 9.8 x(0.0185) }`

v = 13.22 m/s