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 .