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A 7.80-g bullet moving at 660 m/s penetrates a tree trunk to a depth of 6.30 cm?
A 7.80-g bullet moving at 660 m/s penetrates a tree trunk to a depth of 6.30 cm.
(a) Use work and energy considerations to find the average frictional force that stops the bullet.
(b) Assuming the frictional force is constant, determine how much time elapses between the moment the bullet enters the tree and the moment it stops moving.
3 Answers
- FiremanLv 77 years agoFavorite Answer
A 7.80-g bullet moving at 660 m/s penetrates a tree trunk to a depth of 6.30 cm
(a) By work energy relation:-
=>W =∆KE
=>F.s = 1/2mv^2
=>F x 6.30 x 10^-2 = 1/2 x 7.80 x 10^-3 x (660)^2
=>F = 26965.71 N
(b) By F = ma
=>a = 26965.71/(7.80 x 10^-3)
=>a = 3457142.86 m/s^2
=>By v = u - at
=>0 = 660 - 3457142.86 x t
=>t = 0.191 x 10^-3 s OR 0.191 ms
- JamesLv 67 years ago
0.5 mv^2 = F x d
0.5 (.0078)(660^2) = F ( .0630)
F = 0.5 (.0078)(660^2) / ( .0630) N
Change in momentum
mv = Force times time
(.0078)(660) = F x t
t = (.0078)(660) / F
