fisica1-youn-e-freedman-exercicios-resolvidos

7.57: The two design conditions are expressed algebraically as
4
ky = f + mg = 3.66 × 10 N (the condition that the elevator remains at rest when the
spring is compressed a distance y; y will be taken as positive) and
1 2
1
kx 2
mv + mgy − fy = (the condition that the change in energy is the work
2
2
3.66×
10 N
Wother = − fy ). Eliminating y in favor of k by y =
k
leads to
4
4
1 (3.66 × 10
2 k
N)
2
4
4
(1.70 × 10 N)(3.66 × 10 N)
+
k
4
4 (1.96 × 10 N)(3.66 × 10
= 62.5×
10 J +
k
4
N)
.
This is actually not hard to solve for k = 919 N m , and the corresponding x is 39.8 m.
This is a very weak spring constant, and would require a space below the operating range
of the elevator about four floors deep, which is not reasonable. b) At the lowest point, the
spring exerts an upward force of magnitude f + mg . Just before the elevator stops,
however, the friction force is also directed upward, so the net force is
2 f
2
( f + mg)
+ f − mg = 2 f , and the upward acceleration is = 17.0 m s .
m
7.58: One mass rises while the other falls, so the net loss of potential energy is
(0.5000
kg − 0.2000
kg)(9.80
m
s
2
)(0.400
m) = 1.176
J.
This is the sum of the kinetic energies of the animals. If the animals are equidistant from
1 2
the center, they have the same speed, so the kinetic energy of the combination is m
2 totv
,
and
2(1.176 J)
v =
= 1.83 m s.
(0.7000 kg)
7.59: a) The kinetic energy of the potato is the work done by gravity (or the potential
energy lost),
b)
1 2
mv = mgl , or v = 2gl
= 2(9.80 m s )(2.50 m) = 7.00 m s .
2
2
v
T − mg = m = 2mg,
l
so T = 3mg
= 3(0.100 kg)(9.80 m s ) = 2.94 N.
2
2