Question

A 7.8 kg object starting from rest falls through a viscous medium and experiences a resistive force R = −b v, where v is the velocity of the object. The acceleration of gravity is 9.8 m/s 2 . If the object’s speed reaches one-half its terminal speed in 6.56 s, determine the terminal speed.

Answer 1

Answer:

The terminal voltage is $v_T = 90.76\ m/s$

Explanation:

From the question we are given that the Resistive force R is mathematically represented as

                    $R = -bv$

       Where b is a constant  known as drag coefficient

          v is the velocity of the object

The objective of this solution is to obtain the terminal speed and this is mathematically represented as

                          $V_T = \frac{mg}{b}$

 Where  m is the mass

               g acceleration due to gravity

       

As the time changes the velocity of the object change and this can be mathematically represented as

                             $v = v_T[1-e^{bt/m}]$

We are told that the  $v = \frac{1}{2} v_T \ at \ t = 6.56s$

              So substituting this into the above equation we have

                                 $\frac{v_T}{2} = v_T [1-e^{bt/m}]$

                                     $\frac{1}{2} = 1- e^{bt/m}$

                    $e^{-(b*6.56)/m}= 1-\frac{1}{2}$

                              $\frac{b}{m} = \frac{ln(2) }{6.56}$

Substituting  m = 7.8 kg

                           $b= \frac{ln(2)}{6.56} *7.8 =0.8242\ kg/m$

Now substituting this value to get the terminal velocity we have

                        $v_T = \frac{7.8 * 9.8}{0.8422} =90.76\ m/s$