fisica1-youn-e-freedman-exercicios-resolvidos

6.76: a) Equating the work done by the spring to the gain in kinetic energy,
1 2 1 2
kx = mv , so
2
0
2
k
v = x
m
400 N / m
(0.060 m)
0.0300 kg
0
=
=
6.93 m / s.
b) W
tot
must now include friction, so 1 2
1 2
mv = W fx
2
tot
= kx
2 0
−
0
, where f is the magnitude
of the friction force. Then,
k 2 2 f
v = x0
− x0
m m
=
400 N / m
0.0300 kg
(0.06 m)
2
−
2(6.00 N)
(0.06 m)
(0.0300 kg)
= 4.90 m /s.
c) The greatest speed occurs when the acceleration (and the net force) are zero, or
f 6.00 N
kx f x = = = 0.0150 m . To find the speed, the net work is
W
= ,
k 400 N / m
tot
1 2 2
= k x − x ) − f ( x − x)
, so the maximum speed is
2
v
(
0
0
max
k 2 2 2 f
= ( x0
− x ) − ( x0
− x)
m m
=
400N / m
(0.0300kg)
((0.060m)
2
2
− (0.0150m) ) −
2(6.00N)
(0.060m
(0.0300kg)
= 5.20 m/s,
which is larger than the result of part (b) but smaller than the result of part (a).
− 0.0150 m)
6.77: Denote the initial compression of the spring by x and the distance from the initial
2
position by L. Then, the work done by the spring is 1
kx and the work done by friction is
− µ
kmg ( x + L)
; this form takes into account the fact that while the spring is compressed,
the frictional force is still present (see Problem 6.76). The initial and final kinetic
1 2
energies are both zero, so the net work done is zero, and
2
kx = µ
kmg( x + L)
. Solving for
L,
2
2
(1/ 2) kx (1/ 2)(250 N / m)(0.250m)
L = − x =
− (0.250m) = 0.813 m,
2
µ
kmg
(0.30)(2.50 kg)(9.80m /s )
or 0.81 m to two figures. Thus the book moves . 81m + .25 m = 1.06 m , or about 1.1 m.
6.78: The work done by gravity is W = −mgLsin
θ (negative since the cat is moving
g
up), and the work done by the applied force is FL, where F is the magnitude of the
applied force. The total work is
2
W = (100 N)(2.00 m) − (7.00 kg)(9.80 m /s )(2.00 m)sin30°
131.4 J.
tot
=
The cat’s initial kinetic energy is
v
1 2 1
2
mv = (7.00 kg)(2.40 m /s) =
2
1
( K + W )
m
2
2
1
2
=
=
=
2
2(20.2 J + 131.4 J)
(7.00 kg)
20.2 J
6.58 m / s.
, and