fisica1-youn-e-freedman-exercicios-resolvidos

6.94: a) The number of cars is the total power available divided by the power needed per
car,
6
13.4×
10 W
= 177,
3
(2.8×
10 N)(27 m/s)
rounding down to the nearest integer.
b) To accelerate a total mass M at an acceleration a and speed v, the extra power
needed is Mav. To climb a hill of angle α , the extra power needed is Mg sin α v.
These
2
will be nearly the same if a ~ g sin α;
if g sin α ~ g tan α ~ 0.10 m/s , the power is
2
about the same as that needed to accelerate at 0.10 m/s .
6
2
c) (1.10× 10 kg)(9.80 m/s )(0.010)(27 m/s) = 2.9 MW. d) The power per car needed
is that used in part (a), plus that found in part (c) with M being the mass of a single car.
The total number of cars is then
6
6
13.4×
10 W − 2.9×
10 W
= 36,
3
4
2
(2.8×
10 N + (8.2×
10 kg)(9.80 m/s )(0.010))(27 m/s)
rounding to the nearest integer.
3
6.95: a) P = Fv = (53×
10 N)(45 m/s) 2.4 MW.
0
=
5
2
b) P
1
= mav = (9.1×
10 kg)(1.5 m/s )(45 m/s) = 61 MW.
c) Approximating sin α , by tan α , and using the component of gravity down the
incline as mg sin α,
5
2
P2 = ( mg sin α ) v = (9.1×
10 kg)(9.80 m/s )(0.015)(45 m/s) = 6.0 MW.
6.96: a) Along this path, y is constant, and the displacement is parallel to the force, so
2
W
2
(2.00 m)
= αy ∫ xdx = ( 2.50 N/m )(3.00 m) = 15.0 J.
2
b) Since the force has no y-component, no work is done moving in the y-direction.
c) Along this path, y varies with position along the path, given by y = 1.5x,
so
F x
2
= α (1.5 x)
x = 1.5αx
, and
2
W = ∫ Fxdx
= 1.5α
∫ x dx
3
2 (200 m)
= 1.5(2.50 N/m )
3
= 10.0 J.