How Does a Bullet's Kinetic Energy Transform When It Passes Through a Tree?

[jai6638]

Q) A 3 g bullet traveling at a speed of 400m/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?

This is an example prob from my book which says:

Change in KE+Change in Internal Energy=0
Hence, Change in Internal Energy=-Changein KE and they find it to be 180J.

I had two questions:

1) Shouldnt the equation be : Change of internal energy= Change in Ke
or KEi +Ui=KEf+Uf .. I can't seem to understand how the change in KE+ Change in Internal Energy equals 0

2) Also, it has a note which says that if the bullet alone was chosen as a system, then work would be done on it and the heat transfer would occur. How would one formulate an equation for this?

Thanks

 

[Astronuc]

The bullet has mass 0.003 kg. Travelling at 400 m/s, it has some kinetic energy, KE = 1/2 m v². It emerges from the tree at a speed of 200 m/s, a lower kinetic energy.

Clearly the energy was lost by interaction with the tree. What is the difference between the kinetic energy before and after the interaction with the tree. What does one think happened to the tree?

 

[HallsofIvy]

Q) A 3 g bullet traveling at a speed of 400m/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?

This is an example prob from my book which says:

Change in KE+Change in Internal Energy=0
Hence, Change in Internal Energy=-Changein KE and they find it to be 180J.

I had two questions:

1) Shouldnt the equation be : Change of internal energy= Change in Ke
or KEi +Ui=KEf+Uf .. I can't seem to understand how the change in KE+ Change in Internal Energy equals 0

"change in kinetic energy" is negative- the bullet slows down! Of course that lost energy is left in the tree: change in total energy of the "tree-bullet system"= change in kinetic energy (of the bullet)+ change in internal energy (of the tree)= 0 is simply conservation of energy.

2) Also, it has a note which says that if the bullet alone was chosen as a system, then work would be done on it and the heat transfer would occur. How would one formulate an equation for this?
You would have to include the work done by friction on the bullet. It would really work out to the same thing except that change in kinetic energy (negative) would be equal to the work done on the bullet by friction in the tree (also negative). Since the work done on the bullet by the tree is the negative of the work done on the tree by the bullet (which is itself the increase in internal energy it's really:
"tree-bullet system"
change in kinetic energy+ change in internal energy= 0
"bullet only"
change in kinetic energy= -(change in internal enery)

Thanks