fisica1-youn-e-freedman-exercicios-resolvidos

2 2 2
2
7.78: a) a = d x dt = −ω
x, F = ma = −m
ω x
c)
a
x
y
= d
2
y
dt
2
0
x
x
2 2
2
= −ω
y = −ω
y, F = ma = −m
ω y
0
0
2
2 2
b) [ ] ω [ ] ( )
y
1 2
x
+ Fydy
= m
0
xdx + ydy
0
x + y
∫ ∫ ω
U = − F dx ∫ ∫
v
v
x
y
0
0
0
0
0
0
0
y
= m
2
= dx dt = −x
ω sin ω t = −x
ω ( y
= dy dt = + y ω cos ω t = + y ω ( x x
(i) When x = x0
and y = 0 , vx
= 0 and v y
= y 0
ω0
2 2 1 2 2 1 2 2
1 2
K = m(
v + vy
) = my0
ω
0
, U = ω0
mx0
and E = K + U = mω0
(
2
2
2
2
(ii) When x = 0 and y = y , v x ω
0 x
= −
0 0
and v
y
= 0
1 2 2 1 2 2
1 2 2
K = ω ,
and
(
2
0mx0
U = mω0
y0
E = K + U = mω0
x0
+ y0
2
2
2
Note that the total energy is the same.
1 2 2
x
x0
+ y0
0
0
0
0
0
y
0
0
)
)
)
)
7.79: a) The mechanical energy increase of the car is
1
2
6
K − K = (1500 kg)(37 m s) = 1.027 10 J.
2 1 2
×
Let α be the number of gallons of gasoline consumed.
8
6
α(1.3
× 10 J)(0.15) = 1.027 × 10 J
α = 0.053gallons
b) ( 1.00 gallons) α = 19 accelerations