fisica1-youn-e-freedman-exercicios-resolvidos

7.20: As in Example 7.8, K
1
= 0 and U = 0.0250
1
J. For v = 0.20 m
2
s,
K = 0.0040
2
J, so 0.0210 J 1 2
,
J)
U
2
= = kx so x = ±
2(0.0210 0.092 m.
2
5.00 N m
= ± In the absence
of friction, the glider will go through the equilibrium position and pass through
x = −0.092 m with the same speed, on the opposite side of the equilibrium position.
7.21: a) In this situation, U = 2
0 when x = 0,
so K = 0.0250 J and
2
2(0.0250 J)
v = 0.500 m s. b) If v
2
= 2.50 m s,
2
=
0.200 kg
2
2(0.625 J)
K = 1 2) (0.200 kg)(2.50 m s) = 0.625 J = , so x = 0.500 m. Or,
2
( U1
1
=
5.00 N m
because the speed is 5 times that of part (a), the kinetic energy is 25 times that of part (a),
and the initial extension is 5 × 0.100 m = 0.500 m.
7.22: a) The work done by friction is
W = −µ
mg∆x
= −(0.05)
(0.200 kg) (9.80 m s
so
other
k
2
) (0.020 m) = −0.00196 J,
K = 0.00704
2
J and 2(0.00704 J)
v =
0.27 m s.
b) In this case W
other
= −0.0098 J, so
2
=
0.200 kg
2(0.0152 J)
K = 0.0250 J − 0.0098 J 0.0152 J, and v = 0.39 m s.
2
=
2 0.200 kg
=
c) In this case, K = 0 2
, U = 0,
so 2
2
U + W = = 0.0250 J − µ (0.200 kg) (9.80 m s ) × (0.100 m), or µ 0.13.
1 other
0
k
k
=
7.23: a) In this case, K = 625,000
1
J as before, W −17,000
other
= J and
U
2
2
= (1 2) ky + mgy
= (1 2)(1.41×
10
2
5
2
2
N m) ( −1.00 m)
2
+ (2000 kg) (9.80 m s ) ( −1.00)
= 50,900 J.
The kinetic energy is then K
2
= 625,000 J − 50,900 J −17,000 J = 557,100 J ,
corresponding to a speed v = 23.6 m
2
s. b) The elevator is moving down, so the friction
force is up (tending to stop the elevator, which is the idea). The net upward force is then
2
5
− mg + f − kx = −(2000 kg)(9.80 m s ) + 17,000 N − (1.41×
10 N m)( −1.00 m) = 138, 400
for an upward acceleration of 69.2 m s .
2