Physics: Aheavy-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,

A baseball was hit at 45 m/s at an angle of 55 degrees above the horizontal.Explanation:...

Aheavy-duty stapling gun uses a 0.179 kg metal rod

Ответ:

lexxiiee653

16.01.2022

Physics

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

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

where

m = mass = 0.179 kgv = final speed at the instant of contactu = initial speed = 0 since it was initially at resthf = final heightHi = initial heightEf = final compression = 1.35 x 10^{-2} mEi = initial compression = 3.2 x 10^{-2} mk = spring constant = 37107 N/mg = 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

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

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

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

v = 13.22 m/s

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