fisica1-youn-e-freedman-exercicios-resolvidos

6.68: From Eq. (6.7), with x = 1
0 ,
x 2 x
2 3 k 2 b 3 c 4
W = ∫ Fdx =
2 2 2
0 ∫
2
( kx − bx + cx ) dx = x − x + x
0
2 3 4
2
2 3
3 4
= (50.0 N / m) x − (233 N / m ) x + (3000 N / m ) x
2
a) When x = 0.050
2
m , W = 0.115 J , or 0.12 J to two figures. b) When
x2 = −0.050 m, W = 0.173 J , or 0.17 J to two figures. c) It’s easier to stretch the spring;
2
the quadratic − bx term is always in the − x -direction, and so the needed force, and
hence the needed work, will be less when x > 2
0 .
2
2
(0.70 m/s)
6.69: a) T ma = m v = 0.120kg) = 0.147 N , or 0.15 N to two figures. b) At
=
rad
(
R
(0.
40 m)
(2.80 m/s)
the later radius and speed, the tension is ( 0.120kg) = 9.41N
(0. 10 m)
, or 9.4 N to two
figures. c) The surface is frictionless and horizontal, so the net work is the work done by
the cord. For a massless and frictionless cord, this is the same as the work done by the
person, and is equal to the change in the block’s kinetic energy,
2 2
2
2
K
2
− K1
= (1/ 2) m(
v2
− v1
) = (1/ 2)(0.120 kg)((2.80 m /s) − (0.70 m /s) ) = 0.441J . Note
that in this case, the tension cannot be perpendicular to the block’s velocity at all times;
the cord is in the radial direction, and for the radius to change, the block must have some
non-zero component of velocity in the radial direction.
6.70: a) This is similar to Problem 6.64, but here α > 0 (the force is repulsive), and
x < , so the work done is again negative;
2
x 1
⎛ 1 1 ⎞
W = α
⎜ −
⎟ = (2.12×
10
⎝ x1
x2
⎠
= −2.65×
10
Note that x
1
is so large compared to
(6.13)) and solving for v
2
,
2
−26
−17
N ⋅ m
J.
2
x that the term
2
2
((0.200 m
1
−1
2
) − (1.25×
10
9
m
−1
))
1
x
is negligible. Then, using Eq.
−17
2 W
5 2 2( −2.65×
10 J)
v
2
= v1
+ = (3.00×
10 m /s) +
= 2.41×
10
−27
m
(1.67 × 10 kg)
α
b) With K2 = 0,
W = −K1. Using W = − ,
2 5
x 2
m/s.
−26
2
α 2α
2(2.12×
10 N ⋅ m )
−10
x
2
= = =
= 2.82×
10 m.
2
−27
5 2
K1
mv1
(1.67 × 10 kg)(3.00×
10 m/s)
c) The repulsive force has done no net work, so the kinetic energy and hence the speed of
5
the proton have their original values, and the speed is 3.00× 10 m / s .