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