Question

Suppose the sand in a sand box with length 14 feet, width 8 feet, and height 1.5 feet is to be removed. The density of the sand (in pounds per cubic feet) h feet from the bottom is given by δ(h)=2.5−h. Find the work required to completely empty the sandbox.

Answer 1

Answer:

The work required is  $W= 252lb \cdot ft$

Explanation:

  The volume of the sand box is mathematical represented as

                     $dV = L * W * dh$

Substituting 14 feet for L,  8 feet for W into the equation

                           $= 14* 8*dh$

                          $= 112dh$

The force as a result of the sand in the disk is mathematically represented as

              $dF = \rho * dV$

     Substituting (2.5-h)  for $\rho$

              $dF =112(2.5-h) dh$

Now the work that is required to lift the sand from h = 0 to a height of  h=1.5 m is mathematically represented as

                      $dW = 112(2.5 -h)(1.5-h)dh$

Now above is the formula for change in work done in order to obtain the workdone we integrate

                    $W = 112 \int\limits^{1.5}_0 {(2.5 - h)(1.5-h)} \, dh$

                         $=112\int\limits^{1.5}_0 {3.5-4h + h^2} \, dh$  

                        $= 112 [3.75h -2h^2 + \frac{h^3}{3} ]{ {{1.5} \atop {0}} \right.$

                       $= 112 [3.75 (1.5) -2(1.5)^2 + \frac{1.5^3}{3} ]$

                      $= 112 * 2.25$

                     $= 252lb \cdot ft$