Question

An object weighing \(32 \mathrm{lb}\) is released from rest \(50 \mathrm{ft}\) above the surface of a calm lake. Before the object reaches the surface of the lake, the air resistance (in pounds) is given by \(2 v\), where \(v\) is the velocity (in feet per second). After the object passes beneath the surface, the water resistance (in pounds) is given by \(6 v\). Further, the object is then buoyed up by a buoyancy force of \(8 \mathrm{lb}\). Find the velocity of the object 2 sec after it passes beneath the surface of the lake.

Short answer

The velocity of the object 2 seconds after passing beneath the surface of the lake is \(-128 \, ft/s\). The negative sign indicates that the object is moving upward due to the combined effect of buoyancy, water resistance, and the previous downward motion.

Step-by-step solution

Calculate the velocity at which the object strikes the water

First, let's analyze the object falling through the air before entering the water. We have two forces acting on the object: a downward force due to gravity and an upward force due to air resistance. Therefore, the net force acting on the object is given by: Net Force = Weight - Air Resistance \( F_{net} = 32 - 2v \) Here, weight \( (32 \, lb) \) is a constant, so gravity is considered to be constant as well. Therefore, we can use Newton's second law, which states that \( F_{net} = ma \), where \( m \) is the mass of the object and \(a\) is its acceleration. The mass of the object can be calculated as \( m = \frac{32}{g} \) slugs, where \( g \approx 32.2 \, ft/s^2 \) is the acceleration due to gravity. Therefore, \( m \approx 1\,slug \). Now, we can rewrite the equation as: \( 32 - 2v = 1a \) We can solve this equation to find the terminal velocity (\(v_t\)), when the object doesn't accelerate anymore.

Calculate the acceleration in the water

When the object enters the water, there are now three forces acting on it: buoyancy force (\(8\,lb\)), water resistance (\(6v\)), and the gravitational force. The net force in the water can be written as: \( F_{net} = 32 - 6v_{water} - 8 \) Using Newton's second law again, we can relate this to acceleration in the water: \( 32 - 6v_{water} - 8 = 1a \) We can now solve this equation to find the acceleration in water.

Find the final velocity of the object after 2 seconds in the water

Using the initial velocity from the end of Step 1 (\(v_t\)) and the acceleration found in Step 2, we can find the final velocity of the object after 2 seconds in the water using the equation: \( v_{final} = v_{initial} + at \) where \(v_{initial}\) is the object's velocity when it enters the water, \(a\) is the acceleration found in Step 2, and \(t = 2s\). Substitute the values and solve for the final velocity after 2 seconds in the water. Solution:

Calculate the velocity at which the object strikes the water

At terminal velocity, net force goes to zero, and we can rewrite the equation as: \( 32 - 2v_t = 0 \) Solve for terminal velocity, \(v_t\): \( v_t = \frac{32}{2} = 16 \, ft/s \)

Calculate the acceleration in the water

Rewrite the equation for net force in water: \( 32 - 6v_{water} - 8 = 1a \) This simplifies to: \( 24 - 6v_{water} = a \) Using the terminal velocity from Step 1 (\(v_t = 16 \, ft/s\)), we can find the acceleration in the water: \( a = 24 - 6(16) = -72 \, ft/s^2 \)

Find the final velocity of the object after 2 seconds in the water

Use the initial velocity (\(v_t\)), acceleration (\(a\)), and time (\(t\)) to calculate the final velocity (\(v_{final}\)): \( v_{final} = 16 + (-72)(2) = 16 - 144 = -128 \, ft/s \) The velocity of the object 2 seconds after passing beneath the surface of the lake is \(-128 \, ft/s\). The negative sign indicates that the object is moving upward due to the combined effect of buoyancy, water resistance, and the previous downward motion.