fisica1-youn-e-freedman-exercicios-resolvidos

1 2
7.46: a) U − U = mg(
h − 2R)
= mv . From previous considerations, the speed at the
A
top must be at least gR . Thus,
B
1
5
mg ( h − 2R)
> mgR, or h > R.
2
2
b) U − U = ( 2.50)
Rmg K , so
A C
=
v C
C
= (5.00) gR =
2
2
A
2
(5.00)(9.80 m s )(20.0 m) = 31.3 m s.
v
The radial acceleration is a
rad
= C
R
= 49.0 m s . The tangential direction is down, the
normal force at point C is horizontal, there is no friction, so the only downward force is
2
gravity, and a = g 9.80 m s .
tan
=
2
7.47: a) Use work-energy relation to find the kinetic energy of the wood as it enters the
rough bottom: U
1
= K2
gives K
2
= mgy1
= 78.4 J.
Now apply work-energy relation to the motion along the rough bottom:
K + U + W = K + U
1
1
other
2
2
Wother
= W
f
= −µ
kmgs, K2
= U1
= U
2
= 0 ; K1
= 78.
4 J
78.4 J − µ
k
mgs = 0 ; solving for s gives s = 20.0 m.
The wood stops after traveling 20.0 m along the rough bottom.
b) Friction does − 78.4 J of work.