56.

(a) By a force analysis in the style of Chapter 6, we find the normal force N = mg cos θ (where

mg = 267 N) which means f

k

= µ

k

mg cos θ. Thus, Eq. 8-29 yields

∆E

th

= f

k

d = µ

k

mgd cos θ = (0.10)(267)(6.1) cos 20

◦

= 1.5

× 10

2

J .

(b) The potential energy change is ∆U = mg(

−d sin θ) = (267)(−6.1 sin 20

◦

) =

−5.6×10

2

J. The initial

kinetic energy is

K

i

=

1

2

mv

2

i

=

1

2

267 N

9.8 m/s

2



(0.457 m/s)

2

= 2.8 J .

Therefore, using Eq. 8-31 (with W = 0), the final kinetic energy is

K

f

= K

i

− ∆U − ∆E

th

= 2.8

−



−5.6 × 10

2



− 1.5 × 10

2

= 4.1

× 10

2

J .

Consequently, the final speed is v

f

=



2K

f

/m = 5.5 m/s