fisica1-youn-e-freedman-exercicios-resolvidos

7.39: a) At constant speed, the upward force of the three ropes must balance the force,
so the tension in each is one-third of the man’s weight. The tension in the rope is the
2
force he exerts, or (70.0 kg)(9.80 m s ) 3 = 229 N.
b) The man has risen 1.20 m, and so
the increase in his potential energy is (70.0 kg)(9.80 m s )(1.20 m) = 823 J.
In moving up
a given distance, the total length of the rope between the pulleys and the platform
changes by three times this distance, so the length of rope that passes through the man’s
hands is 3 × 1.20 m = 3.60 m,
and ( 229 N)(3.6 m) = 824 J.
2
7.40: First find the acceleration:
2 2
2
v − v0
(3.00 m s)
2
a = =
= 3.75 m s
2( x − x0)
2(1.20 m)
Then, choosing motion in the direction of the more massive block as positive:
F = Mg − mg = ( M + m)
a = Ma + ma
net
M ( g − a)
= m(
g + a)
M
m
g + a (9.80 + 3.75) m s
= =
g − a (9.80 − 3.75) m s
M = 2.24 m
Since M + m = 15.0 kg :
2.24m
+ m = 15.0 kg
m = 4.63 kg
M = 15.
0 kg − 4.63 kg = 10.
4 kg
2
2
= 2.24
7.41: a) K
1
+ U1
+ Wother
= K2
+ U
2
U
1
= U
2
= K2
= 0
W = W
f
= −µ
mgs, with s = 280 ft
other k
=
85.3 m
The work-energy expression gives 1 2
mv − µ mgs = 0
v 1
= 2µ k
gs = 22.4 m s = 50 mph; the driver was speeding.
a) 15 mph over speed limit so $150 ticket.
2
1
k