Force with kinetic energy question

[XxMuDvAyNexX]

[SOLVED] Force with kinetic energy question

A 2.5-g bullet traveling at 350 m/s hits a tree and slows uniformly to a stop while penetrating a distance of 12 cm into the tree's truck. What force was exerted on the bullet in bringing it to a rest?



Not sure on this one. I know to find to the kinetic energy would be KE=1/2mv^2. I'm not sure if that is needed though. And to find force would be F=W/D or F=MA.


So far I only have KE=153.125 using the equation above. I plugged in 1/2(.0025kg)(350m/sec)^2. Now I'm not sure what to do. I want to try to find W since it includes distance. I don't know if there is an equation to find work by using only M,V,and X. Hmm..I'm confused now.

 

[lavalamp]

You know the initial and final velocity, and you know the distance over which the acceleration took place. From these three you can calculate the acceleration with $$v^{2} = u^{2} + 2as$$.

From there you can determine the force exerted on the bullet.

 

[Doc Al]

You're doing fine by calculating the KE. How does work done (FD) relate to the change in KE?

 

[XxMuDvAyNexX]

They are equal aren't they? W=K-Ko

 

[XxMuDvAyNexX]

Ahh I think figured it out. The W would be equal to =-153.125J. Then I would just use F=W/D or F=-153.125 J/.12m and that comes out to be...-1276.04. Is that correct?

 

[lavalamp]

Yep, that's right.

 

[XxMuDvAyNexX]

Thank you again!