fisica1-youn-e-freedman-exercicios-resolvidos

1 2
1 2
7.69: With U = 0 , K = 0 K = mv = U = kx + mgh , and solving for v
2
,
2 1
,
2
2
2
2
kx (1900 N m)(0.045 m)
2
v
2
= + 2gh
=
+ 2(9.80 m s )(1.20 m) =
m
(0.150 kg)
2
1
2
7.01 m s
7.70: a) In this problem, use of algebra avoids the intermediate calculation of the spring
constant k. If the original height is h and the maximum compression of the spring is d,
1 2
then mg ( h + d)
= kd . The speed needed is when the spring is compressed 2d
, and from
2
conservation of energy, mg ( h
= mv
h + d ,
⎛ d ⎞ mg(
h + d)
1 2
mg ⎜h
+ ⎟ − = mv ,
⎝ 2 ⎠ 4 2
which simplifies to
2 ⎛ 3 1 ⎞
v = 2g⎜
h + d ⎟.
⎝ 4 4 ⎠
Insertion of numerical values gives v = 6.14 m s . b) If the spring is compressed a
2 1 2
+ d 2) −
1 )
2
k(
d 2
2
. Substituting for k in terms of
1 2
2mg
distance x,
2
kx = mgx , or x =
k
. Using the expression from part (a) that gives k in
terms of h and d,
2
2
d d
x = (2mg)
= = 0.0210 m.
2mg(
h + d)
h + d
7.71: The first condition, that the maximum height above the release point is h, is
1 2
expressed as
2
kx = mgh . The magnitude of the acceleration is largest when the spring is
compressed to a distance x; at this point the net upward force is kx − mg = ma , so the
second condition is expressed as x = ( m k)(
g + a)
. a) Substituting the second expression
into the first gives
1
2
2
⎛ m ⎞
k⎜
⎟
⎝ k ⎠
( g + a)
2
= mgh,
or
m(
g + a)
k =
2gh
gh
b) Substituting this into the expression for x gives = 2 g a
.
x
+
2
.