fisica1-youn-e-freedman-exercicios-resolvidos

7.49: a) K
1
+ U1
+ Wother
= K2
+ U
2
Let point 1 be point A and point 2 be point B. Take y = 0 at point B.
1 2 1 2
mgy + mv mv , with h = 20.0 m and v = 10.0
1
m s
1 2 1
=
2
2
2
v
2
= v1
+ 2gh
= 22.2 m s
b) Use K1 + U1
+ Wother
= K2
+ U2,
with point 1 at B and point 2 where the spring has
its maximum compression x.
1 2
U = U = K = ; K = mv with v 22.2 m s
W
1 2 2
0
1 2 1
1
=
1 2
other
W
f
+ Wel
= −
kmgs
−
2
kx ,
= µ with s = 100 m + x
The work-energy relation gives K +W 0.
1 2
1 2
2 1
−
=
kmgs
−
2
kx
mv µ
0
1 other
=
2
Putting in the numerical values gives x + 29.4x − 750 = 0.
The positive root to this
equation is x =16.4 m.
b) When the spring is compressed x = 16.4 m the force it exerts on the stone is
F
el
= kx = 32.8 N. The maximum possible static friction force is
max f
s
= µ
smg
= (0.80)(15.0 kg)(9.80 m s ) = 118 N.
The spring force is less than the maximum possible static friction force so the stone
remains at rest.
2
7.50: First get speed at the top of the hill for the block to clear the pit.
1 2
y = gt
2
1
2 2
20 m = (9.8 m s ) t
2
t = 2.0 s
Energy conservation:
v
KE
1
2
Top
Bottom
mv
v
t = 40 m → v
2
B
B
= PE
Top
= mgh +
=
=
v
2
T
(20 m s)
= 42 m s
1
2
Top
+ KE
mv
+ 2gh
2
2
T
=
Top
40 m
20 s
+ 2(9.8 m s
= 20 m s
2
)(70 m)