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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  • $\begingroup$ There's not really a conceptual question here. It's a thinly-disguised drill problem. $\endgroup$ – Bill N Jan 29 '16 at 20:50
  • $\begingroup$ 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. $\endgroup$ – NGST01 Jan 30 '16 at 0:15
  • $\begingroup$ Are you in a class or trying to learn this on your own? $\endgroup$ – Bill N Jan 30 '16 at 5:14
  • $\begingroup$ 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. $\endgroup$ – NGST01 Jan 30 '16 at 7:04
  • $\begingroup$ 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. $\endgroup$ – Bill N Jan 30 '16 at 16:17

The work energy theorem stares that the work done is equal to the change in kinetic energy. Work is force time distance moved in the direction of the force. To do work the force must move. Do not forget that you are finding the average force during the interaction. If you know the time over the interaction takes place then this will give you the average acceleration.

  • $\begingroup$ Sweet, So Question time again. Does Impact force for projectiles apply force to the things they hit? such as a bullet hits wood, the force of energy loss over penetration distance will have the force exerted on the object right? $\endgroup$ – NGST01 Jan 30 '16 at 0:10
  • $\begingroup$ The average force exerted on the bullet by the wood is equal to the kinetic energy lost by the bullet divided by the distance the bullet penetrates the wood. $\endgroup$ – Farcher Jan 30 '16 at 0:58
  • $\begingroup$ Okay, BUT the bullet is exerting that force onto the target as well RIGHT? I know this is a dumb question but with Instant energy loss at a certain point of penetration lets say 200ft*lbs lost at .02ft is 10,000 Newtons of force at .02ft and this in turn is 10,000 Newtons exerted on the target right? That explains things like Temporary Stretch Cavities on bullet impacts right? $\endgroup$ – NGST01 Jan 30 '16 at 5:08
  • $\begingroup$ 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). $\endgroup$ – Bill N Jan 30 '16 at 16:21

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