A small blimp is used for advertising purposes at a football game. It has a mass of \(93.5 \mathrm{~kg}\) and is attached by a towrope to a truck on the ground. The towrope makes an angle of \(53.3^{\circ}\) downward from the horizontal, and the blimp hovers at a constant height of \(19.5 \mathrm{~m}\) above the ground. The truck moves on a straight line for \(840.5 \mathrm{~m}\) on the level surface of the stadium parking lot at a constant velocity of \(8.90 \mathrm{~m} / \mathrm{s}\). If the drag coefficient \(\left(K\right.\) in \(\left.F=K v^{2}\right)\) is \(0.500 \mathrm{~kg} / \mathrm{m}\), how much work is done by the truck in pulling the blimp (assuming there is no wind)?

To find the tension in the towrope, we first need to find the gravitational force acting on the blimp. The gravitational force can be calculated using the formula: \(F_g = m \times g\) Where \(F_g\) is the gravitational force, \(m\) is the mass of the blimp (93.5 kg), and \(g\) is the acceleration due to gravity (approximately \(9.81 \mathrm{~m} / \mathrm{s^2}\)). Now calculate the gravitational force: \(F_g = 93.5 \times 9.81 = 916.35 \mathrm{~N}\) Next, since the blimp is hovering at a constant height, the vertical component of the tension in the towrope must be equal to the gravitational force. So we can use trigonometry to find the tension in the towrope: \(T \sin(53.3) = F_g\) Now solve for T: \(T = \frac{F_g}{\sin(53.3)} = \frac{916.35}{\sin(53.3)} = 1186.62 \mathrm{~N}\)

The force exerted by the truck on the blimp is equal to the horizontal component of the tension in the towrope. Therefore: \(F_{truck} = T \cos(53.3) = 1186.62 \times \cos(53.3) = 780.25 \mathrm{~N}\)

The drag force can be calculated using the drag force equation: \(F_{drag} = K \times v^2\) Where \(F_{drag}\) is the drag force, \(K\) is the drag coefficient (0.500 kg/m), and \(v\) is the velocity of the blimp (8.90 m/s). Now calculate the drag force: \(F_{drag} = 0.5 \times (8.90)^2 = 35.43 \mathrm{~N}\)

The net force on the blimp is the difference between the force exerted by the truck and the drag force: \(F_{net} = F_{truck} - F_{drag} = 780.25 - 35.43 = 744.82 \mathrm{~N}\)

Now we can use the work-energy theorem to calculate the work done by the truck on the blimp: \(W = F_{net} \times d\) Where \(W\) is the work done, \(F_{net}\) is the net force on the blimp (744.82 N), and \(d\) is the distance the truck moved (840.5 m). Calculate the work done: \(W = 744.82 \times 840.5 = 625832.61 \mathrm{~J}\) The work done by the truck in pulling the blimp is approximately \(625,832.61 \mathrm{~J}\) (Joules).