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8.5 Power 255

Equation (8.40) is a special instance of a simple formula, which expresses the
instantaneous power as the scalar product of force and velocity. To see this, consider
that when a body suffers a small displacement ds, the force F acting on the body will
perform an amount of work F

dW F ds (8.41)
or q
ds ds is a small displacement
dW F ds cos in the direction of motion.

where is the angle between the direction of the force and the direction of the dis- FIGURE 8.15 The force F makes an angle
placement (see Fig. 8.15). The instantaneous power delivered by this force is then with the displacement ds.
dW ds
P F cos u (8.42)
dt dt
Since ds dt is the speed v, this expression for the power equals

P Fv cos (8.43)
or P F v (8.44) power delivered by a force

A horse pulls a sled up a steep snow-covered street of slope 1:7
EXAMPLE 9
(see Fig. 8.16a).The sled has a mass of 300 kg, and the coeffi-
cient of sliding friction between the sled and the snow is 0.12. If the horse pulls par-
allel to the surface of the street and delivers a power of 1.0 hp, what is the maximum
(constant) speed with which the horse can pull the sled uphill? What fraction of
the horse’s power is expended against gravity? What fraction against friction?
SOLUTION: Figure 8.16b is a “free-body” diagram for the sled, showing the weight
(w mg), the normal force (N mg cos ), the friction force ( f N ), and
k k
the pull of the horse (T ).With the x axis along the street and the y axis at right angles
to the street, the components of these forces are
w mg sin w mg cos
x y
N 0 N mg cos
x y
f mg cos f 0
k,x k k,y
T T T 0
x y
Since the acceleration along the street is zero (constant speed), the sum of the
x components of these forces must be zero:

mg sin 0 mg cos T 0 (8.45)
k
We can solve this equation for the pull of the horse:

T mg sin mg cos (8.46)
k
This simply says that the pull of the horse must balance the component of the
weight along the street plus the friction force.The direction of this pull is parallel
to the direction of motion of the sled. Hence, in Eq. (8.43), 0, and the power
delivered by the horse is
P Tv (mg sin mg cos )v (8.47)
k
Solving this equation for v, we find