Problem#1
A 9.00-kg object starting from rest falls through a viscous medium and experiences a resistive force R = bv, where v is the velocity of the object. If the object reaches one-half its terminal speed in 5.54 s, (a) determine the terminal speed. (b) At what time is the speed of the object threefourths the terminal speed? (c) How far has the object traveled in the first 5.54 s of motion?Answer:
the speed of an object falling in a viscous object is given by
v = (mg/b)[1 – exp(–bt/m)]
where exp(x) = ex is the exponential function.
At t → ∞; v → vT = mg/b
At t = 5.54 s; v = 0.500vT, then
0.500vT = vT[1 – exp{–b(5.54 s)/9.00 kg}]
0.500 = exp{–b(5.54 s)/9.00 kg}
–b(5.54 s)/9.00 kg = ln(5.00) = –0.693
b = (9.00 kg)(0.693)/5.54 s
b = 1.13 kg/s
so that
(a) the terminal speed is
vT = mg/b = (9.00 kg)(9.80 m/s2)/1.13 m/s = 78.3 m/s
(b) v = vT[1 – exp(–bt/m)]
0.750vT = vT[1 – exp{–(1.13 m/s)t/(9.00 kg)}]
exp{–(1.13 m/s)t/(9.00 kg)} = 0.250
–(1.13 kg/s)t/(9.00 kg) = ln 0.250
t = –9.00 kg (ln 0.250)/1.13 m/s = 11.1 s
(c) dx/dt = (mg/b)[1 – exp(–bt/m)]
dx = (mg/b)[1 – exp(–bt/m)]dt
∫x0x dx = (mg/b)∫0t[1 – exp(–bt/m)]dt
x – x0 = mgt/b + (m2g/b2)exp(–bt/m)│0t
x – x0 = mgt/b + (m2g/b2)[exp(–bt/m) – 1]
at t = 5.54 s,
x – 0 = (9.00 kg)(9.80 m/s2)(5.54 s)/(1.13 kg/s) + (9.00 kg)2(9.80 m/s2)/(1.13 kg/s)2[exp(–1.13 kg/s)(5.54 s)/(9.00 kg) – 1]
x = 434 m + (626 m)(– 0.500) = 121 m
Problem#2
A model airplane of mass 0.750 kg flies in a horizontal circle at the end of a 60.0-m control wire, with a speed of 35.0 m/s. Compute the tension in the wire if it makes a constant angle of 20.0° with the horizontal. The forces exerted on the airplane are the pull of the control wire, the gravitational force, and aerodynamic lift, which acts at 20.0° inward from the vertical as shown in Figure 1.
 |
| Fig.1 |
Answer:
the forces acting on the plane are shown in the Fig. 2 below,
∑Fy = may
Flift cos20.00 – T sin20.00 – mg = 0
Flift cos20.00 – T sin20.00 – (0.750 kg)(9.80 m/s2) = 0
Flift cos20.00 – T sin20.00 = 7.35 N (*)
and
∑Fx = mv2/r
Flift sin20.00 + T cos20.00 = (0.750 kg)(35.0 m/s)2/(60.0 m cos20.00)
Flift sin20.00 + T cos20.00 = 16.3 N (**)
Eq.(*) and (**) becomes
Flift – T tan20.00 = sec20.00(7.35 N)
Flift + T cot20.00 = cosec20.00(16.3 N)
Then
T(cot20.00 – tan20.00) = cosec20.00(16.3 N) – sec20.00(7.35 N)
T(3.11) = 39.8 N
T = 12.8 N
Problem#3
Members of a skydiving club were given the following data to use in planning their jumps. In the table, d is the distance fallen from rest by a sky diver in a “free-fall stable spread position,” versus the time of fall t. (a) Convert the distances in feet into meters. (b) Graph d (in meters) versus t. (c) Determine the value of the terminal speed vT by finding the slope of the straight portion of the curve. Use a least-squares fit to determine this slope.
| t(s) | d(ft) | t(s) | d(ft) |
| 1 | 16 | 11 | 1309 |
| 2 | 62 | 12 | 1483 |
| 3 | 138 | 13 | 1657 |
| 4 | 242 | 14 | 1831 |
| 5 | 366 | 15 | 2005 |
| 6 | 504 | 16 | 2179 |
| 7 | 652 | 17 | 2353 |
| 8 | 808 | 18 | 2527 |
| 9 | 971 | 19 | 2701 |
| 10 | 1138 | 20 | 1875 |
Answer:
(a) Convert the distances in feet into meters
| t(s) | d(m) | t(s) | d(m) |
| 1 | 4.88 | 11 | 399 |
| 2 | 18.9 | 12 | 452 |
| 3 | 42.1 | 13 | 505 |
| 4 | 73.8 | 14 | 558 |
| 5 | 112 | 15 | 611 |
| 6 | 154 | 16 | 664 |
| 7 | 199 | 17 | 717 |
| 8 | 246 | 18 | 770 |
| 9 | 296 | 19 | 823 |
| 10 | 347 | 20 | 876 |
(b) Graph d (in meters) versus t
(c) A straight line fits the points from t = 11 0 . s to 20.0 s quite precisely. Its slope is the terminal
speed.
vT = SLOPE = (876 m – 399 m)/(20.0 s – 11.0 s) = 53.0 m/s
Problem#4
If a single constant force acts on an object that moves on a straight line, the object’s velocity is a linear function of time. The equation v = vi + at gives its velocity v as a function of time, where a is its constant acceleration. What if velocity is instead a linear function of position? Assume that as a particular object moves through a resistive medium, its speed decreases as described by the equation v = vi – kx, where k is a constant coefficient and x is the position of the object. Find the law describing the total force acting on this object.
Answer:
v = vi – kx implies the acceleration is
a = dv/dt
a = dvi/dt – kdx/dt
a = 0 – kv = –kv
Then the total force is
∑F = ma = m(–kv)
The resistive force is opposite to the velocity:
∑F = –kmv
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