fisica1-youn-e-freedman-exercicios-resolvidos

7.6: a) (Denote the top of the ramp as point 2.) In Eq. (7.7),
K
2
= 0,
Wother
= −(35 N) × (2.5 m) = −87.5
J, and taking U = 1
0 and
2
2(147 J+
87.5 J)
U = mgy = 12 kg) (9.80 m s ) (2.5 m sin 30°
) = 147 J, v =
6.25 m s, or
2 2
(
1
=
12 kg
6 .3 m s to two figures. Or, the work done by friction and the change in potential energy
are both proportional to the distance the crate moves up the ramp, and so the initial speed
m
is proportional to the square root of the distance up the ramp; 5.0 m s) 6.25 m s.
K
b) In part a), we calculated W
other
and U
2
. Using Eq. (7.7),
1
2
= (12 kg) (11.0 m s) − 87.5 J −147
J 491.5 J
2
=
2
v
2K2 2(491.5 J)
2
=
m
=
(12 kg)
=
9.05 m s.
2.5
( =
1.6 m
7.7: As in Example 7.7, K
2
= 0,
U2
= 94 J, and U = 0 3
. The work done by friction is
− ( 35 N) (1.6 m) = −56
J, and so K = 38 3
J, and 2(38 J)
v
2.5 m s.
3
= =
12 Kg
7.8: The speed is v and the kinetic energy is 4K. The work done by friction is
proportional to the normal force, and hence the mass, and so each term in Eq. (7.7) is
proportional to the total mass of the crate, and the speed at the bottom is the same for any
mass. The kinetic energy is proportional to the mass, and for the same speed but four
times the mass, the kinetic energy is quadrupled.
7.9: In Eq. (7.7), K
1
= 0,
Wother
is given as − 0.22 J, and taking
U = , K = mgR 0.22 J, so
2
0
2
−
v
⎛
2
0.22 J ⎞
2⎜(9.80 m s ) (0.50 m) − ⎟
⎝
0.20 kg ⎠
2
=
=
2.8 m/s.