fisica1-youn-e-freedman-exercicios-resolvidos

6.42: The initial and final kinetic energies of the brick are both zero, so the net work done
2
on the brick by the spring and gravity is zero, so (1 2) kd − mgh = 0 , or
2
d = 2mgh
/ k = 2(1.80 kg)(9.80 m /s )(3.6 m)/(450 N / m) = 0.53 m. The spring will
provide an upward force while the spring and the brick are in contact. When this force
goes to zero, the spring is at its uncompressed length.
5
6.43: Energy = (power)(time) = (100 W)(3600 s) = 3.6×
10 J
K
6.44: Set time to stop:
1 2
= mv so v = 2K/m
= 100 s for m = 70 kg.
2
ΣF = ma : µ mg = ma
a = µ
kg
v v + at
= 0
k
2
= ( 0.200)(9.80 m / s ) =
1.96 m / s
2
0 = 8.00 m /s − (1.96 m /s )t
t = 4.08 s
1 2
KE 2
mv
P = =
t t
1
2
2
(20.0 kg)(8.00 m / s )
=
= 157 W
4.08 s
2
3
6.45: The total power is (165 N)(9.00 m / s) = 1.485×
10 W , so the power per rider is
742.5 W, or about 1.0 hp (which is a very large output, and cannot be sustained for long
periods).
6.46: a)
19
( 1.0 × 10 J / yr)
11
= 3.2 × 10 W.
7
(3.16×
10 s / yr)
b)
11
3.2 × 10 W
= 1.2 kW/person.
8
2.6×
10 folks
c)
11
3.2×
10 W
8 2
2
= 8.0 × 10 m = 800 km .
3 2
(0.40)1.0 × 10 W / m