Chapter 1: Q103P (page 168)

You throw a rock downward into water with a speed of 3 mg/k, where k is the coefficient in Eq. (5.5). Assume that the relationship between fluid resistance and speed is as given in Eq. (5.5), and calculate the speed of the rock as a function of time.

Short Answer

Expert verified

The speed of the rock as a function of time is $v = \frac{2 m g}{k} [\frac{1}{2} + e^{- (\frac{k t}{m})}]$ .

Step by step solution

Identify the given data

The speed of the rock is $v = \frac{m g}{k}$ .

Find the speed of the rock as a function of time

Using Newton’s second law for the falling ball as:

$m \frac{d v}{d t} = k v_{t} - m g \frac{m}{k} \frac{d v}{d t} = k v_{t} - \frac{m g}{k}$

Here, m is mass of an object, k is constant, $v_{t}$ is terminal velocity, and $v$ is the velocity of the object.

It is given that $\frac{m g}{k} = v$ .

$\frac{m}{k} \frac{d v}{d t} = v_{t} - v - \frac{m}{k} \frac{d v}{d t} = v - v_{t} \frac{d v}{v - v_{t}} = - \frac{k}{m} d t$

Integrate both sides of the above equation by taking the lower limit as $3 v_{t}$ .

$\int_{3 v_{t}}^{v} \frac{d v}{v - v_{t}} = - \frac{k t}{m} \ln {(v - v_{t})}_{3 v_{t}}^{v} = - \frac{k t}{m} \ln \frac{v - v_{t}}{2 v_{t}} = - \frac{k t}{m} \ln [\frac{v}{2 v_{t}} - \frac{1}{2}] = - \frac{k t}{m}$

Simplify further as:

$v = 2 v_{t} [\frac{1}{2} + e^{- (\frac{k t}{m})}] = \frac{2 m g}{k} [\frac{1}{2} + e^{- (\frac{k t}{m})}]$

Therefore, the speed of the rock as a function of time is $v = \frac{2 m g}{k} [\frac{1}{2} + e^{- (\frac{k t}{m})}]$ .

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You throw a rock downward into water with a speed of 3 mg/k, where k is the coefficient in Eq. (5.5). Assume that the relationship between fluid resistance and speed is as given in Eq. (5.5), and calculate the speed of the rock as a function of time.

Short Answer

Step by step solution

Identify the given data

Find the speed of the rock as a function of time

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