Answered

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 :

usmanbiu

Answer:

v = 13.22 m/s

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

[tex]0.5.mv^{2} + mg(hf) + 0.5k(Ef)^{2} = 0.5.mu^{2} + mg(hi) + 0.5k(Ei)^{2}[/tex]

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

[tex]0.5.mv^{2} + mg(hf) + 0.5k(Ef)^{2} = mg(hi) + 0.5k(Ei)^{2}[/tex]

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

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

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

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

v = 13.22 m/s

Other Questions