fisica1-youn-e-freedman-exercicios-resolvidos

7.28: a) From (0, 0) to (0, L), x = 0 and so F r
= 0,
, and the work is zero. From (0, L) to
(L, L), F r and d l r
are perpendicular, so F
r ⋅ d l
r
= 0.
and the net work along this path is
zero. b) From (0, 0) to (L, 0), F
r ⋅ d l
r
= 0.
From (L, 0) to (L, L), the work is that found in
2
2
the example, W2 = CL , so the total work along the path is CL . c) Along the diagonal
path, x = y, and so F
r ⋅ d l
r
2
CL
= Cy dy ; integrating from 0 to L gives . (It is not a
2
coincidence that this is the average to the answers to parts (a) and (b).) d) The work
depends on path, and the field is not conservative.
7.29: a) When the book moves to the left, the friction force is to the right, and the work
is − ( 1.2 N)(3.0 m) = −3.6
J. b) The friction force is now to the left, and the work is again
− 3.6 J. c) − 7.2 J.
d) The net work done by friction for the round trip is not zero, and
friction is not a conservative force.
7.30: The friction force has magnitude µ (0.20)(30.0 kg)(9.80 m/s
2
k
mg =
) = 58.8 N. a)
For each part of the move, friction does − ( 58.8 N)(10.6 m) = −623 J,
so the total work
done by friction is − 1.2 kN.
b) − ( 58.8 N)(15.0 m) = −882 N.
7.31: The magnitude of the friction force on the book is
2
µ
k
mg = (0.25)(1.5 kg)(9.80 m s ) = 3.68 N.
a) The work done during each part of the motion is the same, and the total work done
is − 2(3.68 N)(8.0 m) = −59
J (rounding to two places). b) The magnitude of the
displacement is 2 (8.0 m) , so the work done by friction is
− 2(8.0 m)(3.68 N) = −42 N. c) The work is the same both coming and going,
and the total work done is the same as in part (a), − 59 J.
d) The work required to
go from one point to another is not path independent, and the work required for a
round trip is not zero, so friction is not a conservative force.
1 2 2
2 2
7.32: a) k(
x − x ) b) − 1 k(
x − x ). The total work is zero; the spring force is
2
1
2
2
1
2
conservative c) From x
1
to x
3,
1 2 2
( x
2 2
W = − k x3
−
1
).
2
From x
3
to x 2
, W =
1 k(
x x 2
−
3
).
2
2 2
The net work is − 1 k(
x − x ). This is the same as the result of part (a).
2
2
1
7.33: From Eq. (7.17), the force is
dU
F x
= − = C6
dx
The minus sign means that the force is attractive.
d
dx
⎛
⎜
⎝
1
6
x
⎞
⎟
⎠
6C
= −
7
x
6
.