I just have this h/w question, and it looks like it's really simple, I might just be missing something.
"A 0.22 calibre rifle shoots a bullet of mass 1.8g with a muzzle velocity of 500 m/s. If the barrel is 25cm long, what is the force exerted on the bullet while it is in the barrel?"
I know F=ma, I have the mass but not the acceleration, but i have the distance (25 cm=0.25 m) and the v1=500 m/s .... the answer is said to be 900 N, meaning that the acceleration would have to be 500 m/s^2 ... I dont understand how that makes sense... can anyone explain?
2006-11-04 14:42:38 · 3 answers · asked by ab237 1 in Science & Mathematics ➔ Physics
You're in luck. I teach physics and I know exactly how to answer this question (it's based on the work-energy principle). It's pretty simple, really. But it would be best if we discussed it by messenger. Yahoo Messenger me, my ID is
fortitudinousskeptic
- Kevin
2006-11-04 14:46:37 · answer #1 · answered by Anonymous · 1⤊ 0⤋
You have more information. At the exit, you have a kinetic energy. This is equal to the work exerted by the gas, hence, is related to the force*distance. Solve for force.
btw.. practice safe internet computing...
2006-11-04 14:51:13 · answer #2 · answered by arbiter007 6 · 0⤊ 0⤋
W = change in KE
F*d = 1/2 mv^2
F = (1/2)(0.0018kg)(500m/s)^2 / (0.25 m)
F = 900 N
2006-11-04 14:53:44 · answer #3 · answered by Ghidorah 3 · 0⤊ 0⤋