fisica1-youn-e-freedman-exercicios-resolvidos

7.1: From Eq. (7.2),
mgy =
(800 kg) (9.80 m
2
6
s ) (440 m) = 3.45×
10
J = 3.45
MJ.
7.2: a) For constant speed, the net force is zero, so the required force is the sack’s
2
weight, (5.00 kg)(9.80 m s ) = 49 N. b) The lifting force acts in the same direction as the
sack’s motion, so the work is equal to the weight times the distance,
( 49.00 N) (15.0m) = 735 J; this work becomes potential energy. Note that the result is
independent of the speed, and that an extra figure was kept in part (b) to avoid roundoff
error.
7.3: In Eq. (7.7), taking K 0 (as in Example 6.4) and U = , K = U + .
1 =
0 2 2 1
Wother
Friction does negative work − fy,
so K2 = mgy − fy;
solving for the speed v
2,
v
2( mg −
m
f ) y
2
2((200 kg) (9.80 m s ) − 60 N) (3.00 m)
(200 kg)
2
=
=
=
7.55 m s.
.0 m
7.4: a) The rope makes an angle of ( ) °
3
arcsin = 30 with the vertical. The needed
6.0 m
2
2
horizontal force is then w tan θ = (120 kg) (9.80 m s ) tan 30°
= 679 N, or 6.8× 10 N to
two figures. b) In moving the bag, the rope does no work, so the worker does an amount
of work equal to the change in potential energy,
2
3
(120 kg) (9.80 m s ) (6.0 m) (1− cos 30°
) = 0.95×
10 J. Note that this is not the product
of the result of part (a) and the horizontal displacement; the force needed to keep the bag
in equilibrium varies as the angle is changed.
7.5: a) In the absence of air resistance, Eq. (7.5) is applicable. With y
1
− y2
= 22.0 m,
solving for v
2
gives
2
2
2
v
2
= v1
+ 2g(
y2
− y1)
= (12.0 m s) + 2(9.80 m s )(22.0 m) = 24.0 m s.
b) The result of part (a), and any application of Eq. (7.5), depends only on the
magnitude of the velocities, not the directions, so the speed is again 24 .0 m s. c) The
ball thrown upward would be in the air for a longer time and would be slowed more by
air resistance.