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A small blimp is used for advertising purposes at a football game. It has a mass of 93.5 kg93.5 \mathrm{~kg} and is attached by a towrope to a truck on the ground. The towrope makes an angle of 53.353.3^{\circ} downward from the horizontal, and the blimp hovers at a constant height of 19.5 m19.5 \mathrm{~m} above the ground. The truck moves on a straight line for 840.5 m840.5 \mathrm{~m} on the level surface of the stadium parking lot at a constant velocity of 8.90 m/s8.90 \mathrm{~m} / \mathrm{s}. If the drag coefficient (K(K in F=Kv2F=K v^2 ) is 0.500 kg/m0.500 \mathrm{~kg} / \mathrm{m}, how much work is done by the truck in pulling the blimp (assuming there is no wind)?

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Answered 1 year ago
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Given the following data

m=93.5 kgθ=53.3h=19.5 md=840.5 mv=8.90 msK=0.500 kgm\begin{aligned} m&=93.5\ \mathrm{kg}\\ \theta&=53.3^{\circ}\\ h&=19.5\ \mathrm{m}\\ d&=840.5\ \mathrm{m}\\ v&=8.90\ \mathrm{\dfrac{m}{s}}\\ K&=0.500\ \mathrm{\dfrac{kg}{m}}\\ \end{aligned}

our objective is to calculate the work WW done by the truck when pulling the blip.

We'll do so by applying the formula

W=Fd\begin{equation} W=\vec{F}\cdot\vec{d}\end{equation}

Sketching this problem and analyzing the forces is recommended before doing any calculations.

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