Question

Archimedes’ principle of buoyancy states that an object submerged in a fluid is buoyed up by a force equal to the weight of the fluid that is displaced by the object. A rectangular box of 1×2×31 \times 2 \times 3 feet and weighing 384 pounds is dropped into a 100-foot deep freshwater lake. The box begins to sink with a drag due to the water having a magnitude equal to 1/2 the velocity. Calculate the terminal velocity of the box. Will the box have achieved a velocity of 10 feet per second by the time it reaches the bottom? Assume that the density of water is 62.5 pounds per cubic foot.

Solution

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We are given:

weight: mg=384depth: 100 feetdrag force: 12vdensity of water: 62.5 pounds per cubic foot\begin{align*} \text{weight: } mg&=384\\\\ \text{depth: } &100 \text{ feet}\\\\ \text{drag force: } &\frac{1}{2}v \\\\ \text{density of water: }&62.5 \text{ pounds per cubic foot}\\ \end{align*}

Buoyant force is equal to the weight of 1×2×31\times 2\times 3 feet box of water, which equals 12362.5=375 pounds1 \cdot 2 \cdot 3 \cdot 62.5=\colorbox{Lavender}{375 pounds}

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