Q103P - University Physics with Modern Physics

Question

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.

Step-by-Step Solution

Answer

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

Step 1: Identify the given data

• The speed of the rock is $v=\frac{mg}{k}$.

Step 2: Find the speed of the rock as a function of time

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

 

$$\begin{gathered} m\frac{dv}{dt}=k{v}_t-mg \\ \frac{m}{k}\frac{dv}{dt}=k{v}_t-\frac{mg}{k} \end{gathered}$$ 

 

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 $\frac{mg}{k}=v$.

 

$$\begin{gathered} \frac{m}{k}\frac{dv}{dt}=v_t-v \\ -\frac{m}{k}\frac{dv}{dt}=v-v_t \\ \frac{dv}{v-v_t}=-\frac{k}{m}dt \end{gathered}$$

 

Integrate both sides of the above equation by taking the lower limit as $3v_t$.

 

$$\begin{gathered} {\int }_{3v_t}^v\frac{dv}{v-v_t}=-\frac{kt}{m} \\ \ln {\left(v-v_t\right)}_{3v_t}^v=-\frac{kt}{m} \\ \ln \frac{v-v_t}{2v_t}=-\frac{kt}{m} \\ \ln \left[\frac{v}{2v_t}-\frac{1}{2}\right]=-\frac{kt}{m} \end{gathered}$$ 

 

Simplify further as:

 

$$\begin{gathered} v=2v_t\left[\frac{1}{2}+e^{-\left(\frac{kt}{m}\right)}\right] \\ =\frac{2mg}{k}\left[\frac{1}{2}+e^{-\left(\frac{kt}{m}\right)}\right] \end{gathered}$$ 

 

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