Problem#1
A windmill, such as that in the opening photograph of this chapter, turns in response to a force of high-speed air resistance, R = ½ DρAv2. The power available is P = Rv = ½ Dρπr2v3, where v is the wind speed and we have assumed a circular face for the windmill, of radius r. Take the drag coefficient as D = 1.00 and the density of air from the front endpaper. For a home windmill with r = 1.50 m, calculate the power available if (a) v = 8.00 m/s and (b) v = 24.0 m/s. The power delivered to the generator is limited by the efficiency of the system, which is about 25%. For comparison, a typical home needs about 3 kW of electric power.Answer:
Given: r = 1.50 m, D = 1.00, ρ = 1.20 kg/m3, then by using
P = ½ Dρπr2v3, we get
(a) the power available if v = 8.00 m/s, is
P = ½ (1.00)(1.20 kg/m3)π(1.50 m)2(8.00 m/s)3
P = 2.17 kW
(b) the power available if v = 24.0 m/s, is
Pa/Pb = (va/vb)3
Pb = (vb/va)3Pa = (8.00/24.0)3(2.17 kW) = 58.6 kW
Problem#2
More than 2300 years ago the Greek teacher Aristotle wrote the first book called Physics. Put into more precise terminology, this passage is from the end of its Section Eta:
Let P be the power of an agent causing motion; w, the thing moved; d, the distance covered; and ∆t, the time interval required. Then (1) a power equal to P will in a period of time equal to ∆t move w/2 a distance 2d; or (2) it will move w/2 the given distance d in the time interval ∆t/2. Also, if (3) the given power P moves the given object w a distance d/2 in time interval ∆t/2, then (4) P/2 will move w/2 the given distance d in the given time interval ∆t.
(a) Show that Aristotle’s proportions are included in the equation P∆t = bwd where b is a proportionality constant.
(b) Show that our theory of motion includes this part of Aristotle’s theory as one special case. In particular, describe a situation in which it is true, derive the equation representing Aristotle’s proportions, and determine the proportionality constant.
Answer:
(a) The suggested equation P∆t = bwd implies all of the following cases:
(1) P∆t = b(w/2)(2d)
(2) P(∆t/2) = b(w/2)d
(3) P(∆t) = bw(d/2)
(4) (P/2)∆t = b(w/2)d
These are all of the proportionalities Aristotle lists.
(b) For one example, consider a horizontal force F pushing an object of weight w at constant velocity across a horizontal floor with which the object has coefficient of friction µk.
∑F = ma implies that:
+n – w = 0 and F – µkn = 0
So that
F = µkw
As the object moves a distance d, the agent exerting the force does work
W = Fd cosθ = Fd cos00 = µkwd
and puts out power
P = W/∆t
This yields the equation P∆t = µkwd which represents Aristotle’s theory with b = µk.
Our theory is more general than Aristotle’s. Ours can also describe accelerated motion.
Problem#3
A block of mass 1.6 kg is attached to a horizontal spring that has a force constant of 1.0 x 103 N/m, as shown in Figure 2. The spring is compressed 2.0 cm and is then released from rest. If a constant friction force of 4.0 N retards its motion from the moment it is released, (a) At what position x of the block is its speed a maximum? (b) In the What If? section of this example, we explored the effects of an increased friction force of 10.0 N. At what position of the block does its maximum speed occur in this situation?
(a) So long as the spring force is greater than the friction force, the block will be gaining speed. The block slows down when the friction force becomes the greater. It has maximum speed when
kxa – fk = ma = 0
xa = fk/k = 4.00 N/(1.00 x 103 N/m) = 4.00 mm, the left x = 0.
xa = fk/k = 10.00 N/(1.00 x 103 N/m) = 1.00 mm, the left x = 0.
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