fisica1-youn-e-freedman-exercicios-resolvidos

6.38: a) K = 4.0 J , so v = 2 K m = 2(4.0 J) (2.0 kg) = 2.00 m / s . b) No work is
done between x = 3.0 m and x = 4.0 m , so the speed is the same, 2.00 m/s. c)
K = 3.0 J , so v = 2 K / m = 2(3.0 J) /(2.0 kg) = 1.73 m / s .
2
2
6.39: a) The spring does positive work on the sled and rider; ( 1/ 2) kx = (1/ 2) mv , or
v = x k / m = (0.375 m) (4000 N / m) /(70 kg) =
2 2
the spring is (1/ 2) k(
x 1
− x ) , so the final speed is
2
2.83 m / s
. b) The net work done by
k 2 2 (4000 N / m
2
2
v = ( x1 − x2
) =
((0.375 m) − (0.200 m) ) =
m
(70 kg)
2.40 m / s.
6.40: a) From Eq. (6.14), with dl = Rdφ
,
P
2
0
W = ∫ F cosφ
dl = 2wR
cosφ
dφ
= 2wRsinθ0.
P
∫
1
0
In an equivalent geometric treatment, when F r is horizontal, F
r
⋅ d l
r
= Fdx , and the total
work is F = 2w
times the horizontal distance, in this case (see Fig. 6.20(a)) R sinθ0
,
2w giving the same result. b) The ratio of the forces is = 2cotθ0
θ
w tan θ
.
0
2wRsinθ0 sin θ0
θ0
c) = 2 = 2cot .
wR(1
− cosθ
) (1 − cosθ
) 2
0
0
6.41: a) The initial and final (at the maximum distance) kinetic energy is zero, so the
2
positive work done by the spring, ( 1/ 2) kx , must be the opposite of the negative work
done by gravity, − mgLsinθ
, or
2
2mgLsin
θ 2(0.0900 kg)(9.80 m / s )(1.80 m)sin 40.0°
x =
=
= 5.7 cm.
k
(640 N / m)
At this point the glider is no longer in contact with the spring. b) The intermediate
calculation of the initial compression can be avoided by considering that between the
point 0.80 m from the launch to the maximum distance, gravity does a negative amount
2
of work given by − (0.0900 kg)(9.80 m / s )(1.80 m − 0.80 m)sin 40.0°
= −0.567
J , and so
the kinetic energy of the glider at this point is 0.567 J. At this point the glider is no longer
in contact with the spring.