fisica1-youn-e-freedman-exercicios-resolvidos

6.85: f = .25 mg soW
f
= W = −(0.25
mg)
s,
where s is the length of the rough patch.
W
k
= K
0
tot
− K
tot 2 1
1 2 1 2 1
2
1
= mv0
K
2
2
= mv
2 2
= m(0.45v
2
0
) 0. 2025
2
1 2
( )
K , = mv
1
The work-energy relation gives − ( 0.25mg)
s = (0.2025 −1)
mv
2
The mass divides out and solving gives s = 1.5 m.
0
2
0
6.86: Your friend’s average acceleration is
v − v0 6.00 m/s
a = = = 2.00 m/s 2
t 3.00 s
Since there are no other horizontal forces acting, the force you exert on her is given by
2
F
net
= ma = (65.0 kg)(2.00 m/s ) = 130 N
Her average velocity during your pull is 3.00 m/s, and the distance she travels is thus 9.00
m. The work you do is Fx = ( 130 N)(9.00 m) = 1170 J , and the average power is
therefore 1170 J/3.00 s = 390 W. The work can also be calculated as the change in the
kinetic energy.
2
5
5
6.87: a) (800 kg)(9.80 m/s )(14.0 m) = 1.098×
10 J, or 1.10×
10 J to three figures.
2
5
b) (1/ 2)(800 kg)(18.0 m/s ) = 1.30×
10 J.
5
5
1.10×
10 J + 1.30×
10 J
c) = 3.99 kW.
60 s
6.88:
P = Fv = mav
= m(2α
+ 6βt)(2αt
+ 3βt
2
= m(4α
t + 18αβt
2
= (0.96 N/s) t + (0.43 N/s ) t
2
2
)
+ 18β
t
2 3
)
2
3 3
+ (0.043 N/s ) t .
At t = 400 s, the power output is 13.5 W.
6.89: Let t equal the number of seconds she walks every day. Then,
7
(280 J/s) t + (100 J/s)(86400 s − t)
= 1.1×
10 J. Solving for t, t = 13 ,111 s = 3.6 hours.