fisica1-youn-e-freedman-exercicios-resolvidos

6.10: a) From Eq. (6.6),
1 2
5
⎛ ⎛ 1 m / s ⎞⎞
K = (1600 kg) ⎜(50.0 km/h) ⎜ ⎟⎟
= 1.54×
10 J.
2 ⎝ ⎝ 3.6 km / h ⎠⎠
b) Equation (6.5) gives the explicit dependence of kinetic energy on speed; doubling the
speed of any object increases the kinetic energy by a factor of four.
1 1 m / s 2
3
6.11: For the T-Rex, K = (7000 kg)((4 km / hr) ) = 4.32×
10 J . The person’s
2
3.6 km/hr
3
velocity would be v = 2(4.32×
10 J)/70 kg = 111 . m/s , or about 40 km/h.
6.12: (a) Estimate: v ≈ 1m/ s (walking)
v ≈ 2m/ s (running)
m ≈ 70 kg
Walking:
1 2 1
2
KE = mv = (70 kg)(1m / s) =
2
1 2
Running: KE = (70 kg)(2 m / s) = 140 J
2
(b) Estimate: v ≈ 60 mph = 88 ft / s ≈ 30 m / s
m ≈ 2000 kg
1 2
KE = (2000 kg)(30 m / s) = 9 × 10
(c)
KE =
2
KE = W gravity
= mgh Estimate h ≈ 2 m
(1kg)(9.8 m / s
2
2
)(2 m) ≈ 20 J
5
J
35 J
6.13: Let point 1 be at the bottom of the incline and let point 2 be at the skier.
W = K − K
tot
2
1
1 2
K
1
= mv0
, K
2
= 0
2
Work is done by gravity and friction, so W = W mg
+ W
W mg
W f
= −mg
y − y ) = −mgh
(
2 1
(
kmg
cosα)(
h / sinα)
= −µ
k
tot f
.
= − fs = − µ mgh / tanα
Substituting these expressions into the work-energy theorem and solving for v
0
gives
v
0
= 2gh(1
+ µ
k
/ tanα)