fisica1-youn-e-freedman-exercicios-resolvidos

6.27: a) The friction force is µ
k
mg , which is directed against the car’s motion, so the net
2
work done is − µ k
mgs . The change in kinetic energy is ∆ K = −K1 = −( 1/ 2)
mv0
, and so
2
s = v0 / 2µ
k
g . b) From the result of part (a), the stopping distance is proportional to the
square of the initial speed, and so for an initial speed of 60 km/h,
s = (91.2 m)(60.0 /80.0) 2 = 51.3 m . (This method avoids the intermediate calculation of
µ , which in this case is about 0.279.)
k
6.28: The intermediate calculation of the spring constant may be avoided by using Eq.
(6.9) to see that the work is proportional to the square of the extension; the work needed
4.00 cm
to compress the spring 4.00 cm is (12.0 J) ( ) 21.3 J
3.00 cm
2
= .
6.29: a) The magnitude of the force is proportional to the magnitude of the extension or
compression;
( 160 N)(0.015 m / 0.050 m) = 48 N, (160 N)(0.020 m / 0.050 m) = 64 N.
b) There are many equivalent ways to do the necessary algebra. One way is to note
⎛ 169 N ⎞
1 2
that to stretch the spring the original 0.050 m requires
2
⎜ = (0.050 m) = 4 J
0.050 m
⎟
,
⎝ ⎠
so that stretching 0.015 m requires (4 J)(0.015/ 0.050)
2 = 0.360 J and compressing 0.020
m requires
(4 J)(0.020 / 0.050)
2 = 0.64 J . Another is to find the spring constant
3
k = (160 N) ÷ (0.050 m) = 3.20×
10 N / m , from which ( 1/ 2)(3.20 × 10
3
2
= 0.360 J and (1/ 2)(3.20 × 10 N / m)(0.020 m) = 0.64 J .
3
N / m)(0.015 m)
2
6.30: The work can be found by finding the area under the graph, being careful of the
sign of the force. The area under each triangle is 1/2 base× height .
a) 1 / 2 (8 m)(10 N) = 40 J .
b) 1 / 2 (4 m)(10 N) = + 20 J .
c) 1 / 2 (12 m)(10 N) = 60 J .
6.31: Use the Work-Energy Theorem and the results of Problem 6.30.
(2)(40 J)
a) v = = 2.83 m / s
10 kg
b) At x = 12 m , the 40 Joules of kinetic energy will have been increased by 20 J, so
(2)(60 J)
v = = 3.46 m / s .
10 kg