Answer:
The work done lifting the bucket to the top of the building is 341.25 lbf-ft.
Explanation:
Given that bucket is part of a variable-mass system due to sand losses and such work, measured in [tex]lbf \cdot ft[/tex] is associated to gravitational potential energy by Work-Energy Theorem, the work done by lifting the bucket is represented by the following integral:
[tex]W = \frac{1}{g_{c}} \int\limits^{y_{max}}_{y_{min}} {m(y) \cdot g} \, dy[/tex]
Where:
[tex]m(y)[/tex] - Mass of the bucket as a function of height, measured in pounds.
[tex]g[/tex] - Gravitational constant, measured in feet per square second. ([tex]g = 32.174\,\frac{ft}{s^2}[/tex])
[tex]g_{c}[/tex] - lb-f to lb-m conversion factor, measured in lb-m to lb-f. ([tex]g_{c} = 32.174\,\frac{lbm}{lbf}[/tex])
Since leaking is constant, the mass of the bucket can be modelled by using a first-order polynomial (linear function), that is:
[tex]m(y) = m_{o}+\dot m \cdot y[/tex]
Where:
[tex]m_{o}[/tex] - Initial mass of the bucket, measured in pounds.
[tex]\dot m[/tex] - Leaking rate, measured in pounds per feet.
[tex]y[/tex] - Height of the bucket with respect to bottom, measured in feet.
After replacing the mass and simplifying the integral, the following expression for work is found:
[tex]W = \frac{m_{o}\cdot g}{g_{c}}\int\limits^{y_{max}}_{y_{min}}\, dy + \frac{\dot m \cdot g}{g_{c}}\int\limits^{y_{max}}_{y_{min}} {y} \, dy[/tex]
[tex]W = \frac{m_{o}\cdot g}{g_{c}}\cdot (y_{max}-y_{min}) +\frac{\dot m \cdot g}{2\cdot g_{c}}\cdot (y_{max}-y_{min})^{2}[/tex]
If [tex]m_{o} = 15\,pd[/tex], [tex]g = 32.174\,\frac{ft}{s^{2}}[/tex], [tex]g_{c} = 32.174\,\frac{lbm}{lbf}[/tex], [tex]\dot m = -0.3\,\frac{pd}{ft}[/tex] and [tex]y_{max} - y_{min} = 35\,ft[/tex], the work done lifting the bucket to the top of the building is:
[tex]W = \frac{(15\,pd)\cdot \left(32.174\,\frac{ft}{s^{2}} \right)}{32.174\,\frac{lbm}{lbf} }\cdot (35\,ft) + \frac{\left(-0.3\,\frac{pd}{ft} \right)\cdot \left(32.174\,\frac{ft}{s^{2}} \right)}{2\cdot \left(32.174\,\frac{lbm}{lbf} \right)}\cdot (35\,ft)^{2}[/tex]
[tex]W = 341.25\,lbf\cdot ft[/tex]
The work done lifting the bucket to the top of the building is 341.25 lbf-ft.
