# The equation for average impact force over distance traveled/penetrated equals work done/change in kinetic energy [closed]

So the equation Favg * d = Work or change in kinetic energy, is this true? If a car crashes into a tree by 1 meter with 10000 joules, is the impact force on the car 10000 Newtons? If a bullet penetrates .3048 meters with 500 Joules is 1640.41 Newtons exerted on the target it hits, and assumed the impulse is 1 second, should it accelerate and move an object, if it moves that object over a distance, will it do work? Is this correct?

## closed as off-topic by Bill N, Gert, Kyle Kanos, user36790, ACuriousMind♦Jan 30 '16 at 15:26

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• There's not really a conceptual question here. It's a thinly-disguised drill problem. – Bill N Jan 29 '16 at 20:50
• Well, its my uncertainty about something, I just wish I could ask a physicist to assure if I am right, so Im a bit lost. – NGST01 Jan 30 '16 at 0:15
• Are you in a class or trying to learn this on your own? – Bill N Jan 30 '16 at 5:14
• hehe,Im learning on my own...but some other things, I know a lot about this, however I was just wondering if I was 100% correct on these things...but Im in no main physics class, other classes with this...but Im learning this on my own. – NGST01 Jan 30 '16 at 7:04
• This really isn't a site to confirm basic physics concepts that are easily available in a good textbook. There are many learning resources out there. Schaum's outlines gives a wide variety of basic physics examples and exercises. Giancoli or Serway are fairly comprehensive texts. Hyperphysics is a good summary website, but doesn't have exercises. – Bill N Jan 30 '16 at 16:17

• No, you're mixing the units improperly. F(0.02 ft) = 200 ft.lb $\to$ F=10000 lb force. Yes, that magnitude force is bullet on target and target on bullet. Target will break, bullet might (depends on the material structure of each). – Bill N Jan 30 '16 at 16:21