fisica1-youn-e-freedman-exercicios-resolvidos

mg
7.72: Following the hint, the force constant k is found from w = mg = kd , or k =
d
.
When the fish falls from rest, its gravitational potential energy decreases by mgy; this
1 2 1 mg 2
becomes the potential energy of the spring, which is
2
ky =
2 d
y . Equating these,
1 mg
y
2 = mgy,
or y = 2d.
2 d
2 2 2 2
9.73: a) ∆ a = ω r − ω r = ( ω − ω ) r
rad
=
0
[ ω − ω ][ ω + ω ]
0
⎡ω − ω0
⎤
=
⎢
⎣ t ⎥
⎦
=
0
[(
ω + ω ) t]
[ α] [2 ( θ − θ ) r.
b) From the above,
2
2
∆arad
(85.0 m s − 25.0 m s )
2
αr = =
= 2.00 m s .
2∆θ
2(15.0 rad)
c) Similar to the derivation of part (a),
0
0
r
0
r
∆K
= ω
2
1 2
2
0
1
I − ω
2
I
1
= [ α][2∆θ]
I
2
= Iα∆θ.
d) Using the result of part (c),
∆K
45.0 J − 20.0 J)
I = =
= 0.208 kg ⋅ m
2
α∆θ ((2.00 m s )/(0.250 m))(15.0 rad)
( 2
.