fisica1-youn-e-freedman-exercicios-resolvidos

7.56: a) The energy stored may be found directly from
1 2
625,000 J 51,000 J ( 58,000 J) 6.33 10
5
2
= K1
+ Wother
− mgy2
= − − − = ×
ky
J.
2
b) Denote the upward distance from point 2 by h. The kinetic energy at point 2 and
at the height h are both zero, so the energy found in part (a) is equal to the negative of the
work done by gravity and friction,
− ( mg + f ) h = −((2000 kg)(9.80 m s ) + 17,000 N) h = (36,600 N)
h, so
5
2
6.33×
10 J
h = = 17.3 m. c) The net work done on the elevator between the highest point of
4
3.66×
10 J
the rebound and the point where it next reaches the spring is
4
( mg − f )( h − 3.00 m) = 3.72 × 10 J. Note that on the way down, friction does negative
2(3.72 × 10 J)
work. The speed of the elevator is then = 6.10 m s.
4
2000 kg
d) When the elevator
next comes to rest, the total work done by the spring, friction, and gravity must be the
negative of the kinetic energy K
3
found in part (c), or
4
1 2
4
2
K
3
= 3.72 × 10 J = −(
mg − f ) x3
+ kx3
= −(2,600
N) x3
+ (7.03×
10 N m) x3
.
2
(In this calculation, the value of k was recalculated to obtain better precision.) This is
a quadratic in x
3
, the positive solution to which is
x
3
1
=
2(7.03×
10
×
3
3 2
4
4
[ 2.60 × 10 N + (2.60×
10 N) + 4(7.03×
10 N m)(3.72 × 10 J) ]
= 0.746 m,
4
N
m)
5
4
corresponding to a force of 1.05 × 10 N and a stored energy of 3.91× 10 J . It should be
noted that different ways of rounding the numbers in the intermediate calculations may
give different answers.