fisica1-youn-e-freedman-exercicios-resolvidos

4.5: Of the many ways to do this problem, two are presented here.
Geometric: From the law of cosines, the magnitude of the resultant is
2
2
R = (270 N) + (300 N) + 2(270 N)(300 N)cos 60°
= 494 N.
The angle between the resultant and dog A’s rope (the angle opposite the side
corresponding to the 250-N force in a vector diagram) is then
⎛ sin 120°
(300 N) ⎞
arcsin ⎜
⎟ = 31.7°
.
⎝ ( 494 N)
⎠
Components: Taking the + x -direction to be along dog A’s rope, the components of the
resultant are
= ( 270 N) + (300 N)cos 60°
= 420 N
R
x
R
y
= ( 300 N)sin 60°
= 259.8 N,
2
2
259.8
so R = 420 N) + (259.8 N) = 494 N, θ = arctan ( ) = 31.7°
.
(
420
4.6: a) F F = (9.00 N)cos 120° + (6.00 N)cos ( −126.9°
) = 8.10 N
1x
+
2x
−
1 y
+ F2
y
= (9.00 N)sin 120° + (6.00 N)sin ( −126.9°
) = +
F
2 2
2
2
b) R R x
+ R = (8.10 N) + (3.00 N) = 8.64 N.
=
y
3.00 N.
4.7:
= / = / =
2
a F m (132 N) (60 kg) 2.2 m / s (to two places).
2
4.8: F = ma = (135 kg)(1.40 m/s ) = 189 N.
2
4.9: m = F / a = (48.0 N)/(3.00 m/s ) = 16.00 kg.
4.10: a) The acceleration is
m
= a
F
=
80 .0 N
2
=
0.88 m/s
90.9 kg.
a
2x
2(11.0m)
2
= t
2 =
2 = 0.88 m / s
(5.00 s)
. The mass is then
b) The speed at the end of the first 5.00 seconds is at = 4.4 m/s , and the block on the
frictionless surface will continue to move at this speed, so it will move another
vt = 22.0m in the next 5.00 s.