Question

Chapter 14, Problem 042 A flotation device is in the shape of a right cylinder, with a height of 0.588 m and a face area of 4.19 m2 on top and bottom, and its density is 0.346 times that of fresh water. It is initially held fully submerged in fresh water, with its top face at the water surface. Then it is allowed to ascend gradually until it begins to float. How much work does the buoyant force do on the device during the ascent?

Answer 1

Answer:

The workdone is  $W = 9.28 * 10^{3} J$

Explanation:

From the question we are told that

   The height of the cylinder is  $h = 0.588\ m$

   The face Area is  $A = 4.19 \ m^2$

    The density of the cylinder is $\rho = 0.346 * \rho_w$

     Where $\rho_w$ is the density of freshwater which has a constant value

              $\rho_w = 1000 kg/m^3$

     

Now  

     Let the final height of the device under the water be  $= h_f$

      Let  the initial volume underwater be $= V_n$

     Let the initial height under water be  $= h_i$

      Let the final volume under water be  $= V_f$

According to the rule of floatation

        The weight of the cylinder =  Upward thrust

This is mathematically represented as

          $\rho_c g V_n = \rho_w gV_f$

         $\rho_c A h = \rho A h_f$

So      $\frac{0.346 \rho_w}{\rho_w} = \frac{h_f}{h}$

   =>     $\frac{h_f}{h_c} = 0.346$

Now the work done is mathematically represented as  

          $W = \int\limits^{h_f}_{h} {\rho_w g A (-h)} \, dh$

               $= \rho_w g A [\frac{h^2}{2} ] \left | h_f} \atop {h}} \right.$

              $= \frac{g A \rho}{2} [h^2 - h_f^2]$

              $= \frac{g A \rho}{2} (h^2) [1 - \frac{h_f^2}{h^2} ]$

Substituting values

        $W = \frac{(9.8 ) (4.19) (10^3)}{2} (0.588)^2 (1 - 0.346)$

        $W = 9.28 * 10^{3} J$