fisica1-youn-e-freedman-exercicios-resolvidos

6.5: a) See Exercise 5.37. The needed force is
F
2
µ
kmg
(0.25)(30 kg)(9.80 m / s )
=
=
cos φ − µ sinφ
cos30° − (0.25)sin 30°
k
= 99.2 N,
keeping extra figures. b) Fs cos φ = (99.2 N)(4.50 m) cos 30°
= 386.5 J , again keeping an
extra figure. c) The normal force is mg + F sinφ
, and so the work done by friction is
2
− (4.50 m)(0.25)((30 kg)(9.80 m / s ) + (99.2 N)sin 30°
) = −386.5 J . d) Both the normal
force and gravity act perpendicular to the direction of motion, so neither force does work.
e) The net work done is zero.
6.6: From Eq. (6.2),
Fs cosφ = (180 N)(300 m)cos15.0°
= 5.22 × 10
4
J.
6
6.7: 2Fs cosφ
= 2(1.80×
10
3
9
N)(0.75×
10 m)cos14°
= 2.62 × 10 J,
9
or 2.6 × 10 J to
two places.
6.8: The work you do is:
r
F ⋅ s
r = (( 30N)ˆ i − (40N) ˆ) j ⋅ (( −9.0m)ˆ
i − (3.0m) ˆ) j
= ( 30 N)( −9.0 m) + ( −40 N)( −3.0 m)
= −270 N ⋅ m + 120 N ⋅ m = −150 J
6.9: a) (i) Tension force is always perpendicular to the displacement and does no work.
(ii) Work done by gravity is − mg ( y 2
− y1).
When y
1
= y2
, W
mg
= 0 .
b) (i) Tension does no work.
(ii) Let l be the length of the string. = −mg
( y − y ) 1
= −mg(2l)
= 25.1J
W mg
2
−
The displacement is upward and the gravity force is downward, so it does negative
work.