A spring loaded gun has a spring constant of 4.5 N/m. If a 0.05kg bullet is loaded in the gun and the gun is shot horizonally, then what will be the velocity of the bullet when it leaves the gun?
2007-01-22 06:17:12 · 2 answers · asked by Ha!! 2 in Science & Mathematics ➔ Physics
What's the distance that the spring is compressed? The velocity will depend upon that. Once you know that, then take the amount of energy stored in the spring (U = (1/2)*k*x^2, where k is the spring constant (4.5 N/m) and x is the distance which you have not stated yet. The potential energy of the spring will be transferred to the bullet, giving it an equal amount of potential energy, U=(1/2)*m*v^2, where m is the mass of the bullet, 0.05 kg. So you end up with
(1/2)*k*x^2 = (1/2)*m*v^2
Solve for v:
v = squareroot [(k * x^2) / m]
2007-01-22 06:32:01 · answer #1 · answered by Grizzly B 3 · 0⤊ 0⤋
Good question!
F=kx=ma
v=at
s=x=.5at^2
t=(2x/a)^.5
a=kx/m
t=(2x/(kx/m))=(2m/k)^.5
v=at=
v=(kx/m)(2m/k)^.5)
v=x(2k/m)^.5
How much was the spring compressed? What is x=?
2007-01-22 14:32:04 · answer #2 · answered by Edward 7 · 0⤊ 0⤋