Answer:
C = 1.01
Explanation:
Given that,
Mass, m = 75 kg
The terminal velocity of the mass, 
Area of cross section, 
We need to find the drag coefficient. At terminal velocity, the weight is balanced by the drag on the object. So,
R = W
or
Where
is the density of air = 1.225 kg/m³
C is drag coefficient
So,
So, the drag coefficient is 1.01.
By the work-energy theorem, the total work done on the mass as it swings is
<em>W</em> = ∆<em>K</em> = 1/2 (18 kg) (17 m/s)² = 153 J
No work is done by the tension in the string, since it's directed perpendicular to the mass at every point in the arc. Similarly, the component of the mass's weight <em>mg</em> pointing perpendicular to the arc also performs no work.
If we ignore friction/drag for the moment, the only remaining force is the parallel component of weight, which performs <em>mgh</em> = (176.4 N) <em>h</em> of work, where <em>h</em> is the vertical distance between points A and B.
Now, if <em>w</em> is the amount of work done by friction/air resistance, then
(176.4 N) <em>h</em> - <em>w</em> = 153 J
If you know the starting height <em>h</em>, then you can solve for <em>w</em>.
