fisica1-youn-e-freedman-exercicios-resolvidos

2
4.11: a) During the first 2.00 s, the acceleration of the puck is F / m = 1.563 m/s
(keeping an extra figure). At t = 2.00 s , the speed is at = 3.13m/s
and the position is
at
2
/ 2
= vt / 2 = 3.13 m . b) The acceleration during this period is also
2
1 .563 m/s , and the
2
speed at 7.00 s is 3.13 m/s + (1.563 m/s )(2.00 s) = 6.26 m/s . The position at t = 5.00s
is
x = 3 .13 m + (3.13m/s)(5.00s − 2.00s) = 125 m , and at t = 7.00s
is
12.5 m + (3.13 m/s)(2.00s) + (1/2)(1.563 m/s
or 21.9 m to three places.
2
)(2.00 s)
2
= 21.89 m,
2
4.12: a) = F / m = 140 N / 32.5 kg = 4.31m/s .
a x
1 2
b) With v x
= 0, x = at 215 m .
0
=
2
c) With v = 0, v = a t = 2x
/ t 43.0 m/s
0 x x x
= .
4.13: a) ∑ F r
= 0
b), c), d)
4.14: a) With v 0 ,
v
a x
0 x
=
a
2
6 2
vx
(3.00×
10 m/s)
=
= 2.50×
10
2
2x
2(1.80×
10 m)
x
=
−
14
2
m/s .
x 3.00×
10 s
8
b) m /
−
t = = = 1.20×
10 s . Note that this time is also the distance divided by
14 2
2.50×
10 m / s
6
the average speed.
−31 14 2
−16
c) F = ma = (9.11×
10 kg)(2.50×
10 m/s ) = 2.28×
10 N.
2
2
3
4.15: F = ma = w(
a / g)
= (2400 N)(12 m/s )(9.80 m/s ) = 2.94 × 10 N.
F F F ⎛ 160 ⎞
2
2
4.16: a = = = g = ⎜ ⎟(9.80 m/s ) = 22.0 m/s .
m w/
g w ⎝ 71.2 ⎠
2
4.17: a) m = w/
g = (44.0 N) /(9.80 m/s ) = 4.49 kg b) The mass is the same, 4.49 kg, and
2
the weight is (4.49 kg)(1.81m/s ) = 8.13 N.