fisica1-youn-e-freedman-exercicios-resolvidos

6.35: a) The static friction force would need to be equal in magnitude to the spring force,
(20.0
m)
µ s
mg = kd or µ =
N / m)(0.086
1. 76 , which is quite large. (Keeping extra figures in
s 2
=
(0.100 kg)(9.80 m / s )
the intermediate calculation for d gives a different answer.)
relation
1 2 1
µ
kmgd + kd = mv
2 2
2
1
b) In Example 6.6, the
was obtained, and d was found in terms of the known initial speed v
1
. In this case, the
condition on d is that the static friction force at maximum extension just balances the
spring force, or kd = µ mg . Solving for v and substituting,
v
2
1
2
mg 2
= ( µ
s
+ 2µ
sµ
k
)
k
⎛ (0.10 kg)(9.80 m / s
=
⎜
⎝ (20.0 N / m)
from which v = 0.67 m /
1
s .
s
=
=
k
m
k
m
d
2
+ 2gdµ
d
⎛ µ
smg
⎜
⎝ k
2
⎞
⎟
⎠
k
2
1
⎛ µ
smg
⎞
+ 2µ
k
g⎜
⎟
⎝ k ⎠
2
)
2
⎞
⎟((0.60)
⎠
2
+ 2(0.60)(0.47)),
6.36: a) The spring is pushing on the block in its direction of motion, so the work is
positive, and equal to the work done in compressing the spring. From either Eq. (6.9) or
1 2 1
2
Eq. (6.10), W = kx = (200 N / m)(0.025 m) = 0.06 J
2 2
.
b) The work-energy theorem gives
v =
2W
m
=
2(0.06 J)
(4.0 kg)
=
0.18 m / s.
6.37: The work done in any interval is the area under the curve, easily calculated when
the areas are unions of triangles and rectangles. a) The area under the trapezoid is
4 .0 N ⋅ m = 4.0 J . b) No force is applied in this interval, so the work done is zero. c)
The area of the triangle is 1 .0 N ⋅ m = 1.0 J , and since the curve is below the axis
( F
x
< 0) , the work is negative, or − 1.0 J . d) The net work is the sum of the results of
parts (a), (b) and (c), 3.0 J. (e) + 1.0 J − 2. 0 J = −1.0
J .