# Force with kinetic energy question

[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.

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

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

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?

Yep, that's right.

Thank you again!