To solve this problem we will apply the concept of drag force which is described as half the product between the density, the drag coefficient, the area and the squared speed. Said expression mathematically is equivalent to,
[tex]F = \frac{1}{2} \rho C_d Av^2[/tex]
Here,
[tex]\rho[/tex] = Density
[tex]C_d[/tex] = Drag coefficient
A = Area
v = Velocity
Our values are,
[tex]A = 1.8*1.8 = 3.24m^2[/tex]
[tex]C_d = 1.4[/tex]
[tex]v = 6m/s[/tex]
[tex]\rho = 1.23kg/m^3 \rightarrow \text{Density of Air}[/tex]
Replacing at the previous equation we have that,
[tex]F = \frac{1}{2} (1.23)(1.4)(3.24)(6)^2[/tex]
[tex]F = 100.42N[/tex]
Energy can be described through the work theorem, which is the product between force and distance traveled. So,
[tex]W = Fd[/tex]
[tex]W = (100.42)(200)[/tex]
[tex]W= 20084J[/tex]
Therefore the thermal energy is 20.084kJ