fisica1-youn-e-freedman-exercicios-resolvidos

6.32: The work you do with your changing force is
∫
6.9
6.9
F(
x)
dx = ( 20.0 N) dx
0 ∫ − −
0 ∫
6.9
0
N
3.0
m
xdx
6.9 N 2 6.9
= ( −20.0 N) x |
0
−(3.0
)( x / 2) |
0
m
= −138 N ⋅ m − 71.4 N ⋅ m = −209.4 J or − 209 J
The work is negative because the cow continues to advance as you vainly attempt to push
her backward.
6.33: Wtot
= K
2
− K1
1 2
K
1
= mv0
, K
2
= 0
2
1 2
Work is done by the spring force. Wtot
= − kx
2
, where x is the amount the spring is
compressed.
1 2 1 2
− kx = − mv0
and x = v0
m / k = 8.5 cm
2 2
6.34: a) The average force is ( 80.0 J) /(0.200 m) = 400 N , and the force needed to hold
the platform in place is twice this, or 800 N. b) From Eq. (6.9), doubling the distance
quadruples the work so an extra 240 J of work must be done. The maximum force is
quadrupled, 1600 N.
Both parts may of course be done by solving for the spring constant
2
3
k = 2(80.0 J) ÷ (0.
200 m) = 4.
00×
10 N/m , giving the same results.