Q 49 - Calculus

Question

Find the work described in each of Exercises 49–60.

 The work required to lift a full cylindrical soda fountain cup 1 foot into the air, given that the cup is full and has a radius of 3 inches and a height of 6 inches. You may assume that soda has the same weight-density as water.

Step-by-Step Solution

Answer

 The work required to lift a full cylindrical soda = 6.126 foot-pounds

Step 1: Given Information

 The work required to lift a full cylindrical soda fountain cup 1 foot into the air, given that the cup is full and has a radius of 3 inches and a height of 6 inches.

Step 2: Formulas

Volume of cylinder =${\mathrm{\pi r}}^2\mathrm{h}$

where 'r' is the radius and 'h' is the height 

Work required to lift and object is $W=Fd$

F is the object weight 

d is the distance 

Step 3: Convert inches to feet

$$\begin{gathered} 1 in= \frac{1}{12}ft \\ 3 in= \frac{3}{12}=\frac{1}{4}ft \\ 6 in=\frac{6}{12}=\frac{1}{2}ft \\ Volume = {\mathrm{\pi r}}^2\mathrm{h} \\ \mathrm{V}=\mathrm{\pi }{\left(\frac{1}{4}\right)}^2\left(\frac{1}{2}\right) \\ \mathrm{V}=\frac{\mathrm{\pi }}{32}{\mathrm{ft}}^3 \end{gathered}$$

Step 4: Force

Force is given by the formula 

$$\begin{gathered} F=V\omega \\ weight density \omega =62.4 pounds per cubic feet \\ V is the volume \\ F= \frac{\mathrm{\pi }}{32}(62.4) \\ F=6.1261 \end{gathered}$$

Step 5: Work

Now we find out the work required to lift a full cylindrical soda

$$\begin{gathered} W=Fd \\ dis\tan ce =1 \\ F=6.1261 \\ W=6.1261\left(1\right) \\ W=6.1261 \end{gathered}$$

 the work required to lift a full cylindrical soda = 6.126 foot-pounds