fisica1-youn-e-freedman-exercicios-resolvidos

7.62: a) The skier’s kinetic energy at the bottom can be found from the potential energy
at the top minus the work done by friction,
K = mgh −W
= (60.0 kg)(9.8 N kg)(65.0 m) 10.500 J, or
1 F
−
2K
2(27,7 20 J )
K = 38,200 J −10,500
J 27,720 J . Then v =
30.4 m s .
1
=
1
=
m
60 kg
=
b) K + K − W + W ) = 27,720 J − ( µ mgd + f d),
= 27,720 J − [(.2)(588 N)
×
2 1
(
F A
k
air
K
2
(160 N)(82 m)], K
2
= 27,720 J − 22,763 J =
( 82 m) + , or 4957 J . Then,
2K
2(4957 J)
v
2
= =
= 12.85 m s ≈ 12.9
m 60 kg
m s.
F
c) Use the Work-Energy Theorem to find the force. W = ∆KE,
= KE d = ( 4957 J) (2.5 m) = 1983 N ≈ 2000 N.
7.63: The skier is subject to both gravity and a normal force; it is the normal force that
causes her to go in a circle, and when she leaves the hill, the normal force vanishes. The
vanishing of the normal force is the condition that determines when she will leave the
hill. As the normal force approaches zero, the necessary (inward) radial force is the radial
2
component of gravity, or mv R = mg cosα,
where R is the radius of the snowball. The
speed is found from conservation of energy; at an angle α , she has descended a vertical
1 2
2
distance R ( 1−
cosα)
, so mv = mgR(1
− cosα)
, or v = 2gR(1
− cosα)
. Using this in
2
⎛ 2 ⎞
the previous relation gives 2 (1 − cos α)
= cos α , or α = arccos⎜
⎟ = 48. 2°
. This result
⎝ 3 ⎠
does not depend on the skier’s mass, the radius of the snowball, or g.
7.64: If the speed of the rock at the top is v t , then conservation of energy gives the
speed v
b
from 1 2 1 2
mv = mv + mg(2
R ) , R being the radius of the circle, and so
2
b
2
t
2 2
mvt
vb = vt
+ 4gR
. The tension at the top and bottom are found from Tt + mg =
R
and
T
2
mvb
b
mg =
R
− , so T T
m 2 2
− = ( v − v ) + 2mg
= 6mg
6 w .
b t R b t
=
2