A bullet with a velocity of 575m/s Use work and energy considerations

AI Thread Summary
A 7.8g bullet traveling at 575m/s penetrates a tree trunk to a depth of 5.50cm, prompting a discussion on calculating the average frictional force and the time it takes to stop. The initial kinetic energy is calculated, leading to a frictional force of approximately 2.34e4 N. For time calculation, the assumption of constant velocity was incorrect; instead, the relationship between impulse and force-time was applied. This corrected approach yielded a time of 1.91e-4 seconds for the bullet to stop. The discussion highlights the importance of understanding deceleration and momentum in such physics problems.
sarahjt1
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Homework Statement



A 7.8g bullet moving at 575m/s penetrates a tree trunk to depth of 5.50cm. (a) Use work and energy considerations to find the average frictional force that stops the bullet. (b) Assuming that the frictional force is constant, determine hot much time elapses between the moment the bullet enters the tree and the moment it stops.

Homework Equations



I figure part (a) by:

KEi - Fk = KEf
KEi= kinestic energy initial, Fk=frictional force, and KEf= kinetic energy final

For part (b) I thought that the t=D/r would be sufficient but it's not?? I don't really understand what other equations to use in this scenario... HELP!

The Attempt at a Solution



For part (a)


1/2mV^2 - Fk*d = 1/2mVf^2
1/2(.0078)(575)^2 - .055Fk = 0
1289.44 - .055Fk = 0
-.005Fk = -1289.44
Fk = 2.34e4 N

For part (b)

t= .055/575
t= 9.5e-4 --> which is not correct, book shows 1.91e-4, huh? :cry:

Homework Statement

 
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What you are doing is assuming that the bullet travels at a constant velocity and suddenly stops after penetrating 5.50cm into the wood o.0, which definitely does not hold true especially when it is decelerating at a constant rate!
Think along the lines about the relation between impulse and force-time.
 
Ohhh... so this is a momentum question.

F delta t = mVi - mVf

so...

F delta t = (.0078)(575)- (.0078)(0)
2.34e4 (delta t)= 4.49
delta t = .000191 = 1.91e-4

Ah.. Thank you! I would have never of seen that relationship... grrr.
 
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