fisica1-youn-e-freedman-exercicios-resolvidos

7.42: a) Equating the potential energy stored in the spring to the block's kinetic energy,
1 2 1 2
kx = mv , or
2
2
v
k
= x
m
=
400 N m
(0.220 m) = 3.11 m s.
2.00 kg
b) Using energy methods directly, the initial potential energy of the spring is the final
1 2
gravitational potential energy, kx = mgLsinθ
, or
2
1 2
1
2
2
kx
2
(400 N m)(0.220 m)
L = =
= 0.821m.
2
mg sinθ
(2.00 kg)(9.80 m s )sin 37.0°
7.43: The initial and final kinetic energies are both zero, so the work done by the spring
1 2
is the negative of the work done by friction, or kx = µ mgl, where l is the distance the
block moves. Solving for µ k
,
(1/ 2) kx
µ =
mgl
2
2
2
(1/ 2)(100 N m)(0.20 m)
=
2
(0.50 kg)(9.80 m s )(1.00 m)
k
=
k
0.41.
7.44: Work done by friction against the crate brings it to a halt:
f x = potential energy of compressed spring
k
360 J
fk
= = 64.29 N
5.60 m
The friction force working over a 2.00-m distance does work
f
kx
= ( −64.29 N)(2.00 m) = −128.6 J. The kinetic energy of the crate at this point is thus
360 J − 128.6 J = 231.4 J, and its speed is found from
mv
2
v
2
2
= 231.4 J
2(231.4 J)
= = 9.256 m
50.
0 kg
v = 3.04 m s
2
s
2
2
7.45: a) mgh = (0.650 kg)(9.80 m s )(2.50 m) = 15.9 J
b) The second height is 0 .75(2.50 m) = 1.875 m,
so second mgh = 11.9 J ; loses
15 .9 J − 11.9 J = 4.0 J on first bounce. This energy is converted to thermal energy.
a) The third height is 0 .75(1.875 m) = 1.40 m,
, so third mgh = 8.9 J ; loses
11 .9 J − 8.9 J = 3.0 J on second bounce.