fisica1-youn-e-freedman-exercicios-resolvidos

6.56:
Let
M = total mass and T = time for one revolution
1
KE = ∫ ( dm)
v
2
M
dm = dx
L
2πx
v =
T
KE =
L
∫
0
1 ⎛ M ⎞ ⎛ 4π
= ⎜ ⎟ 2
2 L
⎜
⎝ ⎠ ⎝ T
2
1 ⎛ M ⎞ ⎛ 2πx
⎞
⎜ dx⎟
⎜ ⎟
2 ⎝ L ⎠ ⎝ T ⎠
2
1 ⎛ M ⎞ ⎛ 4π
⎞
= ⎜ ⎟ 2
2 L
⎜
T
⎟
⎝ ⎠ ⎝ ⎠
2
L
∫
0
x
2
2
dx
3
⎞ ⎛ L ⎞ 2
⎟
⎜ = π
3
⎟
⎠ ⎝ ⎠ 3
2
ML
2
T
2
5 revolutions in 3 seconds →T = 3 5 s
KE = π
3
2 2
2
2
(12.0 kg) (2.00 m)
(3 5 s)
= 877 J.
2
6.57: a) ( 140 N) (3.80 m) = 532 J b) (20.0 kg) (9.80 m s ) (3.80 m) ( − sin 25°
) = −315
J
c) The normal force does no work.
d)
= − f s = −µ
ns = −µ
mgs cos θ
W f
k
k
k
2
= −(0.30)
(20.0 kg) (9.80 m s ) (3.80 m) cos 25°
= −203 J
e) 532 J − 315 J − 203 J = 15 J (14.7 J to three figures).
f) The result of part (e) is the kinetic energy at the top of the ramp, so the speed is
v = 2 K m = 2(14.7 J) (20.0 kg) = 1.21m s.