fisica1-youn-e-freedman-exercicios-resolvidos

6.18: As the example explains, the boats have the same kinetic energy K at the finish line,
2
2
2 2
so 1/ 2) m v = (1/ 2) m v , or, with m = m , v = 2v
. a) Solving for the ratio of the
(
A A
B B
B
2
A A B
speeds, v
A
/ v B
= 2 . b) The boats are said to start from rest, so the elapsed time is the
distance divided by the average speed. The ratio of the average speeds is the same as the
ratio of the final speeds, so the ratio of the elapsed times is t / t v / v = 2 .
B
A
=
A B
6.19: a) From Eq. (6.5), K
2
= K 1
/ 16 , and from Eq. (6.6), W = −( 15/16) K1
. b) No;
kinetic energies depend on the magnitudes of velocities only.
6.20: From Equations (6.1), (6.5) and (6.6), and solving for F,
F
=
∆K
s
=
1
2
m(
v
2
2
s
− v
2
1
)
=
1
2
2
2
(8.00 kg)((6.00 m / s) − (4.00 m / s) )
= 32.0 N.
(2.50 m)
1
2
2
∆K
(0.420 kg)((6.00 m /s) − (2.00 m /s) )
2
6.21: s = =
= 16.8 cm
F (40.0 N)
6.22: a) If there is no work done by friction, the final kinetic energy is the work done by
the applied force, and solving for the speed,
v =
2W
=
m
2Fs
=
m
b) The net work is Fs − f s = F − mg)
s , so
v =
=
k
( µ
k
2( F − µ
kmg)
s
m
2(36.0 N)(1.20 m)
(4.30 kg)
= 4.48 m /s.
2
2(36.0 N − (0.30)(4.30 kg)(9.80 m / s ))(1.20 m)
(4.30 kg)
= 3.61m / s.
(Note that even though the coefficient of friction is known to only two places, the
difference of the forces is still known to three places.)