45. Equation 8-31 provides ∆E

th

=

−∆E

mec

for the energy “lost” in the sense of this problem. Thus,

∆E

th

=

1

2

m

v

2

i

− v

2

f



+ mg (y

i

− y

f

)

=

1

2

(60)(24

2

− 22

2

) + (60)(9.8)(14)

=

1.1

× 10

4

J .

That the angle of 25

◦

is nowhere used in this calculation is indicative of the fact that energy is a scalar

quantity.