Chapter 13 Solutions - Mosinee School District

Chapter 13
1
2
0 0 60 J
2 mv , so 2 60 J
v
50 10
-3
kg
49 m s
13.10 (a)
k
F
x
max
max
230 N
0.400 m
575 N m
(b)
work done PEs
kx
2 2
1
2
1 2
575 N m 0.400 46.0 J
13.11 From conservation of mechanical energy,
KE PE g PE s KE PE f
g PE s or 1
0 mghf
0 0 0 kx i
2
i
2
giving
k
2 mgh
x
2
i
f
2 0.100 kg 9.80 m s2
0.600 m
2
2.00 10
2
m
2.94 103
N m
13.12 Conservation of mechanical energy, ( KE PEg PEs ) f ( KE PEg PE s)
i
, gives
1 2 1
mv 0 0 0 0
2
2 i
2 f
kx , or
v
i
k
m
x
i
5.00 106
N m 3.16 10
2 m 2.23 m s
1000 kg
13.13 An unknown quantity of mechanical energy is converted into internal energy during the collision. Thus, we apply
conservation of momentum from just before to just after the collision and obtain mv i + M(0) = (M + m)V, or the
speed of the block and embedded bullet just after collision is
V
m
M m
v
i
10.0 10
3
kg
2.01 kg
300 m s 1.49 m s
Now, we use conservation of mechanical energy from just after collision until the block comes to rest. This gives
1 2 1
0
2
2 f 2
kx M m V , or
x
f
V
M m
k
2.01 kg
1.49 m s 0.477 m
19.6 N m
Page 13.4