fisica1-youn-e-freedman-exercicios-resolvidos

6.71: The velocity and acceleration as functions of time are
dx
2
v(
t)
= = 2αt
+ 3βt
,
dt
a(
t)
= 2α
+ 6βt
a)
2
3
2
v ( t = 4.00 s) = 2(0.20 m /s )(4.00s) + 3(0.02 m /s )(4.00s) = 2.56 m / s.
b)
2
3
ma = (6.00 kg)(2(0.20 m /s ) + 6(0.02 m /s )(4.00 s) = 5.28 N.
2
c) W = K − K = K = (1/ 2)(6.00 kg)(256 m /s) 19.7 J.
2 1 2
=
6.72: In Eq. (6.14), dl = dx and φ = 31. 0°
is constant, and so
=
W =
∫
P 2
P1
(5.00 N / m
F cosφdl
=
2
∫
)cos31.0°
x2
x1
∫
F cosφdx
1.50 m
1.00 m
2
x dx = 3.39 J.
The final speed of the object is then
2 2W
2 2(3.39 J)
v
2
= v1
+ = (4.00 m /s) + = 6.57 m /s.
m
(0.250 kg)
2 2
6.73: a) K − K = (1/ 2) m(
v − )
2 1
2
v1
b) The work done by gravity is
so the work done by the rider is
= (1/ 2)(80.0 kg)((1.50 m /s)
2
2
− (5.00 m /s) ) = −910 J.
2 3
− mgh = − (80. 0 kg)(9. 80m/ s )(5. 20 m) = − 4. 08× 10 J ,
3 3
−910 J − ( − 4. 08× 10 J) = 3. 17× 10 J .
∞ b b
b
6.74: a) W = ∫ dx =
= .
x n
n−1
n−1
0 x ( −n
−1))
x ( n −1)
x
∞
x0
Note that for this part, for n > 1,
x
1−n
→ 0 as x → ∞ . b) When 0 < n < 1 , the improper
integral must be used,
⎡ b n−1
n−1
⎤
W = lim ⎢ ( x2
− x0
) ,
x2
→∞ ( 1)
⎥
⎣ n −
⎦
n−1
and because the exponent on the x
2
is positive, the limit does not exist, and the integral
diverges. This is interpreted as the force F doing an infinite amount of work, even though
F → 0 as x → .
2
∞
0
6.75: Setting the (negative) work done by the spring to the needed (negative) change in
1 2 1 2
kinetic energy, kx = mv , and solving for the spring constant,
2
2
0
2
2
mv 1200 kg)(0.65 m /s)
k = 0 =
= 1.03 10
2
2
(0.070 m)
×
x
( 5
N / m.