fisica1-youn-e-freedman-exercicios-resolvidos

6.62: a)
W f
= − f s = −µ
mg cos θs
k
= −(0.31)
k
(5.00 kg) (9.80 m
2
s ) cos12.0°
(1.50 m) = −22.3 J
2
(keeping an extra figure) b) (5.00 kg) (9.80 m s ) sin 12.0°
(1.50 m) = 15.3 J.
c) The normal force does no work. d) 15.3 J − 22.3 J = −7.0
J.
2
e) K = K + W = (1 2) (5.00 kg) (2.2 m s) − 7.0 J 5.1J,
and so
2 1
=
2
= 2(5.1J) /(5.00 kg) = 1.4 m / s
v .
6.63: See Problem 6.62: The work done is negative, and is proportional to the distance s
that the package slides along the ramp, W = mg(sin
θ − µ
k
cosθ)
s . Setting this equal to
the (negative) change in kinetic energy and solving for s gives
2
2
(1/ 2) mv1
v1
s = −
=
mg(sinθ
− µ cosθ)
2g(sinθ
− µ cosθ)
As a check of the result of Problem 6.62,
6.64: a) From Eq. (6.7),
k
2
(2.2 m/s)
=
= 2.6 m.
2
2(9.80 m /s )(sin12° − (0.31)cos12°)
k
( 2.2 m / s) 1− (1.5 m) /(2.6 m) = 1.4 m / s .
2
x2
x2
dx ⎡ 1⎤
⎛ 1 1
∫ ∫
⎟ ⎞
W = Fxdx
= − k = − k
⎢ −
⎥
= k
⎜ − .
x
2
1
x1
x ⎣ x⎦
x1
⎝ x2
x1
⎠
1 1
The force is given to be attractive, so F < 0 , and k must be positive. If x > x < ,
x
x
2 1, x2
x1
and W < 0 . b) Taking “slowly” to be constant speed, the net force on the object is zero,
so the force applied by the hand is opposite F
x
, and the work done is negative of that
1 1
found in part (a), or k( − )
x1
x
, which is positive if x
2
2
> x1
. c) The answers have the same
magnitude but opposite signs; this is to be expected, in that the net work done is zero.
6.65:
F = mg( R r
2
E
/ )
2
2
R ⎛ mgR ⎞
W = −
∞
1
∞
⎝ r ⎠
W = K − K K 0
E
∫ = −∫ ⎜ E ⎟
2
R E
Fds
dr = −mgRE
( −(1/
r)
| ) =
2
, tot 2 1 1
=
12
This gives K
2
= mgRE
= 1.25×
10 J
1 2
K
2
= mv2
so v 2
= 2K
2
/ m = 11,000 m / s
2
mgR
E