fisica1-youn-e-freedman-exercicios-resolvidos

7.76: The kinetic energy K′ after moving up the ramp the distance s will be the energy
initially stored in the spring, plus the (negative) work done by gravity and friction, or
1 2
K′
= kx − mg(sinα + µ
k
cosα)
s.
2
Minimizing the speed is equivalent to minimizing K′ , and differentiating the above
d K′
expression with respect to α and setting = 0 gives
dα
0 = −mgs
(cosα − µ
k
sinα),
1
or tan α = , ⎛ 1
µ
⎟ ⎞
α = arctan
k
⎜ . Pushing the box straight up ( α = 90°
) maximizes the
⎝ µ
k ⎠
vertical displacement h, but not s = h sinα
.
7.77: Let x = 0.18
1
m , x = 0.71 m
2
. The spring constants (assumed identical) are then
known in terms of the unknown weight w, 4 kx
1
= w . The speed of the brother at a given
height h above the point of maximum compression is then found from
1 2 1 ⎛ w ⎞
(4k)
x
2
2
= ⎜ ⎟ v + mgh,
2 2 ⎝ g ⎠
or
v
2
2
(4k)
g 2
⎛ x ⎞
2
= x2
− 2gh
= g
⎜ − 2h
⎟,
w
⎝ x1
⎠
2
2
so v = (9.80 m s ) ((0.71 m) (0.18 m) − 2(0.90 m)) = 3.13 m s , or 3 .1 m s to two
figures. b) Setting v = 0 and solving for h,
2 2
2kx2 x2
h = = = 1.40 m,
mg 2x
or 1.4 m to two figures. c) No; the distance x
1
will be different, and the ratio
0.53 m
( ) 2
2
2
x2 ( x1
0.53 m)
x1
x
= x
1
1
1+
x1
= + will be different. Note that on a small planet, with lower g,
x will be smaller and h will be larger.
1
1