Problem#1
A fire helicopter carries a 620-kg bucket at the end of a cable 20.0 m long as in Figure 1. As the helicopter flies to a fire at a constant speed of 40.0 m/s, the cable makes an angle of 40.0° with respect to the vertical. The bucket presents a cross-sectional area of 3.80 m2 in a plane perpendicular to the air moving past it. Determine the drag coefficient assuming that the resistive force is proportional to the square of the bucket’s speed.
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| Fig.1 |
Answer:
∑Fy = may
T cos40.00 – mg = 0
T = mg/cos40.00
T = (620 kg)(9.80 m/s2)/cos40.00 = 7,930 N
and
∑Fx = max
T sin40.00 – R = 0
R = T sin40.00 = (7,930 N) sin40.00 = 5,100 N
We use
R = ½ DρAv2
5,100 N = ½ D(1.20 kg/m3)(3.80 m2)(40.0 m/s)2
D = 1.40
Problem#2
A small, spherical bead of mass 3.00 g is released from rest at t = 0 in a bottle of liquid shampoo. The terminal speed is observed to be vT = 2.00 cm/s. Find (a) the value of the constant b in Equation 6.2, (b) the time , at which the bead reaches 0.632vT, and (c) the value of the resistive force when the bead reaches terminal speed.
Answer:
(a) At terminal velocity,
R = vTb = mg
b = mg/vT = (3.00 x 10-3 kg)(9.80 m/s2)/(2.00 x 10-2 m/s) = 1.47 N.s/m
(b) In the equation describing the time variation of the velocity, we have
v = vT(1 – e-bt/m)
0.632vT = vT(1 – e-bt/m)
e-bt/m = 0.368
or at time
t = –(m/b) ln (e-bt/m)
t = –(3.00 x 10-3 kg/1.47 N.s/m) ln (0.368)
t = 2.04 x 10-3 s
(c) At terminal velocity,
R = vTb = mg
R = (3.00 x 10-3 kg)(9.80 m/s2) = 2.94 x 10-2 N
Problem#3
The mass of a sports car is 1200 kg. The shape of the body is such that the aerodynamic drag coefficient is 0.250 and the frontal area is 2.20 m2. Neglecting all other sources of friction, calculate the initial acceleration of the car if it has been traveling at 100 km/h and is now shifted into neutral
and allowed to coast.
Answer:
The resistive force is
R = ½ DρAv2
= ½ (0.250)(1.20 kg/m3)(2.20 m2)(27.8 m/s)2
R = 255 N
Then
–R = ma
a = –R/m = –255 N/1200 kg = –0.212 m/s2
Problem#4
A motorboat cuts its engine when its speed is 10.0 m/s and coasts to rest. The equation describing the motion of the motorboat during this period is v = vie-ct, where v is the speed at time t, vi is the initial speed, and c is a constant. At t = 20.0 s, the speed is 5.00 m/s. (a) Find the constant c. (b) What is the speed at t = 40.0 s? (c) Differentiate the expression for v(t) and thus show that the acceleration of the boat is proportional to the speed at any time.
Answer:
(a) the constant c, we use
v = vie-ct
v(20.0 s) = 5.00 = vie-20.0c, because vi = 10.0 m/s
5.00 = 10.0e-20.0c
½ = e-20.0c
–20.0c = ln(½)
c = 3.47 x 10-2/s
(b) at t = 40.s s,
v = (10.0 m/s)e-40.0c = (10.0 m/s)(0.250) = 2.50 m/s
(c) the acceleration of the boat is proportional to the speed at any time given by
a = dv/dt
with v = vie-ct
a = –cvie-ct = –cv
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