fisica1-youn-e-freedman-exercicios-resolvidos

7.60: a) The change in total energy is the work done by the air,
⎛ 1 2 2 ⎞
( K2
+ U
2)
− ( K1
+ U1)
= m⎜
( v2
− v1
) + gy2
⎟
⎝ 2
⎠
2
2
⎛(1
2) ((18.6 m s) − (30.0 m s)
= (0.145 kg) ⎜
2
2
⎝ − (40.0 m s) ) + (9.80 m s )(53.6
= −80.0
J.
⎞
⎟
m)
⎠
b) Similarly,
( K
3
+ U ) − ( K + U
3
2
2
2
2
⎛(1
2)((11.9 m s) + ( −28.7
m s) ⎞
) = (0.145 kg) ⎜
⎟
2
2
(18.6 m s) ) (9.80 m s )(53.6 m)
⎝ −
−
⎠
= −31.3 J.
c) The ball is moving slower on the way down, and does not go as far (in the x-direction),
and so the work done by the air is smaller in magnitude.
7.61: a) For a friction force f, the total work done sliding down the pole is mgd − fd .
This is given as being equal to mgh, and solving for f gives
( d − h)
⎛ h ⎞
f = mg = mg⎜1
− ⎟.
d ⎝ d ⎠
When h = d, f = 0 , as expected, and when h = 0 , f = mg ; there is no net force on the
2 1.0 m
fireman. b) ( 75 kg)(9.80 m s )(1 − ) = 441 N
2.5 m
. c) The net work done is
( mg − f )( d − y)
, and this must be equal to
1 2
mv
2
. Using the above expression for f,
1 2
mv
2
= ( mg −
f )( d − y)
⎛ h ⎞
= mg⎜
⎟(
d − y)
⎝ d ⎠
⎛ y ⎞
= mgh⎜1
− ⎟,
⎝ d ⎠
from which v = 2gh(1
− y d ) . When y = 0 v = 2gh
, which is the original condition.
When y = d , v = 0 ; the fireman is at the top of the pole.