# Homework Help: First Law of Thermodynamics and a bullet

1. Nov 8, 2013

### soccer4life

1. A 3.0-g bullet traveling at a speed of 400 m/s enters a tree and exits the other side with a speed of 200 m/s. Where did the bullet's lost KE go, and what was the energy transferred?

2. ΔKE+ΔPE+ΔU = Q-W

3. The only questions I have about this problem are why does the Work=0, Heat=0, and why does the ΔPE=0?

Last edited by a moderator: Nov 8, 2013
2. Nov 8, 2013

### Staff: Mentor

Welcome to the PF.

Why do you say that work & delta heat are zero? The lost KE has to go somewhere.

And how do you define the PE of the bullet in this problem...?

3. Nov 8, 2013

### soccer4life

Well, this is an example from the book, and the book states in it's solution:
'Q=W=ΔPE=0.' I don't understand why this is true.

Tell me if this is right: ΔPE=0 because the change of height before & after the bullet passes through the tree is so small that it is assumed to =0.

4. Nov 8, 2013

### Staff: Mentor

I have no idea what they are saying with work and heat delta being zero. Work is done by the tree on the bullet to slow it down and change its KE, and the bullet hole is definiely left warmed by the passing of the bullet. Can you scan the example and upload it as an attachment?

5. Nov 8, 2013

### soccer4life

Rather than scan the example, I will copy it's reasoning:

"APPROACH: Take the Bullet and the tree as our system. No Potential energy is involved. No work is done on (or by) the system by outside forces, nor is any heat added because no energy was transferred to or from the system due to a temperature difference. Thus the kinetic energy gets transformed into internal energy of the bullet and tree."

I am confused because I also thought that the tree is doing work on the bullet.

Is this a valid explanation for Q=0: no heat is added or removed from the bullet itself

6. Nov 8, 2013

### haruspex

It all depends how the variables are defined. If Q and W are defined as external inputs/outputs to/from a system consisting of tree+bullet then, yes, they're 0. The heat then becomes represented by the internal energy of the system, U.