fisica1-youn-e-freedman-exercicios-resolvidos

4.33: a) The resultant must have no y-component, and so the child must push with a force
with y-component ( 140 N)sin30° − (100 N)sin 60°
= −16.6
N. For the child to exert the
smallest possible force, that force will have no x-component, so the smallest possible
force has magnitude 16.6 N and is at an angle of 270 ° , or 90 ° clockwise from the
+ x -direction.
b)
∑ F 100 N cos 60°+
140 N cos 30°
2
m = =
= 85.6 kg. w = mg = (85.6 kg)(9.80 m / s ) 840 N.
.
a
2 =
2.0 m / s
4.34: The ship would go a distance
2
2
2
7
2
v0 v0
mv0
(3.6 × 10 kg)(1.5 m /s)
= = =
= 506.25 m,
4
2a
2( F / m)
2F
2(8.0 × 10 N)
so the ship would hit the reef. The speed when the tanker hits the reef is also found from
so the oil should be safe.
4
2
2 2(8.0 × 10 N)(500 m)
v = v − ( 2Fx
/ m)
= (1.5 m/s) −
7
(3.6 × 10 kg)
0
=
0.17 m/s,
2 2
4.35: a) Motion after he leaves the floor: v = v + a ( y − y ).
v = 0 at the maximum height,
y
v
0 y
= 4.85 m/s.
2
b) a
av
= ∆v
/ ∆t
= (4.85 m / s) /(0.300 s) = 16.2 m/s .
c)
y
0 y
2
y 0
2
y − y0 = 1.2 m, a
y
= −9.80
m/s , so
F − w =
av
ma av
F
av
= w + maav
= 890 N +
3
F = 2.36 10 N
av
×
(890 N / 9.80 m / s
2
)(16.2 m / s
2
)
4.36:
F
2
2
v0
(12.5 m / s)
ma = m = (850 kg)
2
2x
2(1.8 × 10 m)
=
−
= 3.7 × 10
6
N.