Answered: An object with a mass of 0.32 kg is…

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Question

An object with a mass of 0.32 kg is released from rest while immersed within a fluid. It reaches a speed of 49.3% its maximum speed in a time of 0.344 seconds. What is the object's terminal speed if the resistive force acting is given by;

R = -bv

Expert Solution

Step 1

Given:

The mass of object is $m = 0.32 kg$ .
The initial speed of object is $u = 0 m / s$ .
The final speed of object is $v = 49.3 % v_{m}$ .
The time taken to reach final speed is $t = 0.344 s$ .

The formula to calculate the net force acting on the object is,

$F = m g + R$

Here, F is the net force, m is the mass, g is the acceleration due to gravity and R is the resistive force.

The resistive force acting on the object is given by,

$R = - b v$

Here, b is the constant and v is the speed.

Substitute the known value in the formula to obtain the net force.

$F = m g - b v$

Step 2

When the object resistive force is equal to the gravitational force then the net force acting on the object is zero and that speed of object is the maximum speed or terminal speed.

$\begin{array}{rcl} 0 & = & m g - b v_{m} \\ b v_{m} & = & m g \\ v_{m} & = & \frac{m g}{b} \end{array}$

When the object accelerates then the formula to calculate the force on the object is,

$F = m a$

Here, a is the acceleration.

Substitute the known value in the equation of force and solve.

$\begin{array}{rcl} m a & = & m g - b v \\ m \frac{d v}{d t} & = & m g - b v \\ m d v & = & (m g - b v) d t \\ \frac{m d v}{m g - b v} & = & d t \end{array}$

Step 3

Substitute $m g - b v = x$ which implies,

$\begin{array}{rcl} - b d v & = & d x \\ d v & = & - \frac{d x}{b} \end{array}$

Substitute this value in the equation and integrate the equation.

$\begin{array}{rcl} \int \frac{m (- \frac{d x}{b})}{x} & = & \int d t \\ - \frac{m}{b} \int \frac{d x}{x} & = & t \\ - \frac{m}{b} \ln x & = & t + C \end{array}$

Here, C is the constant of integration.

Again substitute the original value of $x$ in the equation.

$- \frac{m}{b} \ln (m g - b v) = t + C$

When $t = 0$ then $v = 0$ . Substitute these values in above equation and solve to obtain the value of constant.

$\begin{array}{rcl} - \frac{m}{b} \ln (m g - b (0)) & = & 0 + C \\ C & = & - \frac{m}{b} \ln (m g) \end{array}$

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