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 
  
 
COMENTÁRIO
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
  
DA PROVA – FÍSICA
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




 


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 
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
 
 
 
 
 
COMENTÁRIO NA
F
A
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ASSUNTO:
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MECÂNICA
dA = dB
NB
mA
m1
=
3
(2L )3
L
B
FB,A
PB
PA
I
FA,B = mB
Como, FA,B = FB,A, substituindo II
F – mBa = mA . a
F = mAa + mB . a
F = 9mAa
8mA
Item: D
  
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
FB, A

mA
mB
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L3
F – FB,A = mA . a
  
Logo,
II
em I teremos:
dA = dB
9mA . a
F
=
FA,B
mA . a
9mA a
F
9
=
=
FA,B 8mA a
8
=
8L3
mB = 8mA