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256 CHAPTER 8 Conservation of Energy

(a)
Horse’s pull
is parallel to
inclined surface.

f

(b)
y N
x axis is chosen
x parallel to inclined
street.
T
O f k For constant velocity,
net F and net F y
x
f are zero.
w

(c)
y N
N y
x
T x
f k,x
T
O f k w y = –mg cos f
FIGURE 8.16 (a) Horse Components of
weight perpendicular
dragging a sled up a street. f and parallel to street.
w
(b) “Free-body” diagram for
the sled. (c) Components of w x = –mg sin f
the forces.

P P
v (8.48)
mg sin f m mg cos f mg (sin f m cos f)
k k
For a slope of 1:7, the tangent of the angle of inclination is tan 1/7, and, using
a calculator, the inverse tangent of 1/7 gives 8.1 . Hence

746 W
v
2
300 kg 9.81 m/s (sin 8.1 0.12 cos 8.1 )
0.98 m/s
The weight of the sled makes an angle of 90.0 8.1 98.1 with the direction
of motion (see Fig. 8.16b).The power exerted by the weight of the sled is given by
Eq. (8.43), with F mg and cos cos 98.1 :

2
P mgv cos 98.1 300 kg 9.81 m/s 0.98 m/s cos 98.1
weight
406 W 0.54 hp
Since the total power is 1.0 hp, this says that 54% of the horse’s power is
expended against gravity and, consequently, the remaining 46% against friction.
The friction portion can also be calculated directly.The friction force acts opposite
to the velocity (cos 1), and so the power exerted is negative:

2
P f v mg cos v 0.12 300 kg 9.81 m/s cos 8.1 0.98 m/s
friction k k
343 W 0.46 hp