fisica1-youn-e-freedman-exercicios-resolvidos

6.23: a) On the way up, gravity is opposed to the direction of motion, and so
2
W = −mgs
= −(0.145 kg)(9.80 m /s )(20.0 m) = −28.4 J .
2 W
2 2( −28.4 J)
b) v
2
= v1
+ 2 = (25.0 m /s) + = 15.26m / s .
m
(0.145 kg)
c) No; in the absence of air resistance, the ball will have the same speed on the way
down as on the way up. On the way down, gravity will have done both negative and
positive work on the ball, but the net work will be the same.
6.24: a) Gravity acts in the same direction as the watermelon’s motion, so Eq. (6.1) gives
W
= Fs = mgs
=
(4.80 kg)(9.80 m / s
2
)(25.0 m) = 1176 J.
b) Since the melon is released from rest, K = 1
0 , and Eq. (6.6) gives
K = K
2
= W = 1176 J.
6.25: a) Combining Equations (6.5) and (6.6) and solving for v
2
algebraically,
2 Wtot
2 2(10.0 N)(3.0 m)
v
2
= v1
+ 2 = (4.00 m / s) +
=
m
(7.00 kg)
4.96 m / s.
Keeping extra figures in the intermediate calculations, the acceleration is
2
2
a = (10.0 kg ⋅ m /s ) /(7.00 kg) = 1.429 m /s . From Eq. (2.13), with appropriate change in
notation,
2 2
2
2
v
2
= v1
+ 2as
= (4.00 m / s) + 2(1.429 m / s )(3.0 m),
giving the same result.
6.26: The normal force does no work. The work-energy theorem, along with Eq. (6.5),
gives
2K
2W
v = = = 2gh
=
m m
2gL
sin θ ,
where h = Lsinθ
is the vertical distance the block has dropped, and θ is the angle the
plane makes with the horizontal. Using the given numbers,
v =
2(9.80 m / s
2
)(0.75 m) sin 36.9°
=
2.97 m / s.