43.
(a) We take the initial gravitational potential energy to be U
i
= 0. Then the final gravitational potential
energy is U
f
=
−mgL, where L is the length of the tree. The change is
U
f
− U
i
=
−mgL = −(25kg)
9.8 m/s
2
(12 m) =
−2.9 × 10
3
J .
(b) The kinetic energy is
K =
1
2
mv
2
=
1
2
(25kg)(5.6 m/s)
2
= 3.9
× 10
2
J .
(c) The changes in the mechanical and thermal energies must sum to zero. The change in thermal
energy is ∆E
th
= f L, where f is the magnitude of the average frictional force; therefore,
f =
−
∆K + ∆U
L
=
−
3.92
× 10
2
J
− 2.94 × 10
3
J
12 m
= 210 N .