Problem#1
A sky diver of mass 80.0 kg jumps from a slow-moving aircraft and reaches a terminal speed of 50.0 m/s. (a) What is the acceleration of the sky diver when her speed is 30.0 m/s? What is the drag force on the diver when her speed is (b) 50.0 m/s? (c) 30.0 m/s?
Answer:
Given: m = 80.0 kg, vT = 50.0 m/s, we use
mg = DρAvT2/2, or
DρA/2 = mg/vT2 = (80.0 kg)(9.80 m/s2)/(50.0 m/s2) = 0.314 kg/s
(a) the acceleration of the sky diver when her speed is 30.0 m/s is
a = g – [DρAv2/2]/m
a = 9.80 m/s2 – [(0.314 kg/s)(30.0 m/s)2/(80.0 kg)] = 6.27 m/s2 downward
(b) At v = 50.0 m/s , terminal velocity has been reached.
∑F = 0 = mg – R
R = mg = (80.0 kg)(9.80 m/s2) = 784 N directed up
(c) At v = 30.0 m/s
DρAv2/2 = (0.314 kg/s)(30.0 m/s)2 = 283 N upward
Problem#2
A small piece of Styrofoam packing material is dropped from a height of 2.00 m above the ground. Until it reaches terminal speed, the magnitude of its acceleration is given by a = g – bv. After falling 0.500 m, the Styrofoam effectively reaches terminal speed, and then takes 5.00 s more to reach the ground. (a) What is the value of the constant b? (b) What is the acceleration at t = 0? (c) What is the acceleration when the speed is 0.150 m/s?
Answer:
(a) acceleration is given by a = g – bv
When v = vT, a = 0 and g = bvT and b = g/vT
The Styrofoam falls 1.50 m at constant speed vT in 5.00 s.
Thus, vT = y/t = 1.50 m/5.00 s = 0.300 m/s
Then, b = (9.80 m/s2)/(0.300 m/s) = 32.7 s-1
(b) at t = 0, v = 0 and a = g = 9.80 m/s2 down
(c) When v = 0.150 m/s,
a = g – bv
a = (9.80 m/s2) – (32.7 s-1)(0.150 m/s) = 4.90 m/s down
Problem#3
(a) Estimate the terminal speed of a wooden sphere (density 0.830 g/cm3) falling through air if its radius is 8.00 cm and its drag coefficient is 0.500. (b) From what height would a freely falling object reach this speed in the absence of air resistance?
Answer:
(a) ρ = m/V and A = 0.0201 m2,
and m = ρbeadV = (0.830 g/cm3)[4π(8.00 cm)2/3] = 1.78 kg
Assuming a drag coefficient of D = 0.500 . for this spherical object, and taking the density of air at 20°C from the endpapers, we have
R = ½ ρairADvT2 = mg
½ (1.20 kg/m3)(0.0201 m2)(0.500)vT2 = (1.78 kg)(9.80 m/s2)
0.00603vT2 = 17.444
vT2 = 2,892.87
vT = 53.8 m/s
(b) given by
vf2 = vi2 + 2gh
(53.8 m/s)2 = 0 + 2(9.80 m/s2)h
h = 148 m
Problem#4
Calculate the force required to pull a copper ball of radius 2.00 cm upward through a fluid at the constant speed 9.00 cm/s. Take the drag force to be proportional to the speed, with proportionality constant 0.950 kg/s. Ignore the buoyant force.
Answer:
Since the upward velocity is constant, the resultant force on the ball is zero. Thus, the upward applied force equals the sum of the gravitational and drag forces (both downward):
F = mg + bv
The mass of the copper ball is
m = 4πρr3/3 = 4π(8,920 kg/m3)(0.020 m)3 = 0.299 kg
The applied force is then
F = mg + bv = (0.299 kg)(9.80 m/s2) + (0.950 kg/s)(0.090 m/s)
F = 3.01 N
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