fisica1-youn-e-freedman-exercicios-resolvidos

6.82: The total work done is the sum of that done by gravity (on the hanging block) and
that done by friction (on the block on the table). The work done by gravity is (6.00 kg) gh
and the work done by friction is − µ k
(8.00 kg) gh,
so
2
W
tot
= (6.00 kg − (0.25)(8.00 kg) (9.80 m s ) (1.50 m) = 58.8 J.
This work increases the kinetic energy of both blocks;
so
v =
1 2
W
tot
= ( m1
+ m2)
v
2
2(58.8 J)
(14.00 kg)
,
= 2.90 m s.
6.83: See Problem 6.82. Gravity does positive work, while friction does negative work.
Setting the net (negative) work equal to the (negative) change in kinetic energy,
1
2
( m
1
− µ
km2
) gh = − ( m1
+ m2)
v ,
2
and solving for µ
k
gives
2
m1
+ (1 2) ( m1
+ m2
) v gh
µ
k
=
m
(6.00 kg) + (1 2) (14.00 kg) (0.900 m s)
=
(8.00 kg)
= 0.79.
2
2
2
((9.80 m s ) (2.00 m))
6.84: The arrow will acquire the energy that was used in drawing the bow (i.e., the work
done by the archer), which will be the area under the curve that represents the force as a
function of distance. One possible way of estimating this work is to approximate the F vs.
x curve as a parabola which goes to zero at x = 0 and x = x 0
, and has a maximum of F
0
x 0
2
4F
(
2
x 0
0
0
at x = , so that F x)
= x(
x − x).
This may seem like a crude approximation to the
figure, but it has the ultimate advantage of being easy to integrate;
x
x
0 0
2 3
4F0
2 4F
⎛
0
x0
x ⎞
0
2
∫ Fdx = ( x0x
x ) dx x0
F0
x0.
2
2
x
∫ − =
=
2 3 3
0
0
x
⎜ −
⎟
0
0 ⎝ ⎠
With F = 200 0
N and x 0.75 m, 100
0
= W = J. The speed of the arrow is then
W 2(100 J)
= 89 m s. Other ways of finding the area under the curve in Fig. (6.28)
2
=
m (0.025 kg)
should give similar results.