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A bucket begins weighing 15 pounds, including the sand it holds. The bucket is to be lifted to the top of a 35 foot tall building by a rope of negligible weight. However, the bucket has a hole in it, and leaks 0.3 pounds of sand each foot it is lifted. Find the work done lifting the bucket to the top of the building.

User BorisMoore
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1 Answer

6 votes

Answer:

The work done lifting the bucket to the top of the building is 341.25 lbf-ft.

Step-by-step explanation:

Given that bucket is part of a variable-mass system due to sand losses and such work, measured in
lbf \cdot ft is associated to gravitational potential energy by Work-Energy Theorem, the work done by lifting the bucket is represented by the following integral:


W = (1)/(g_(c)) \int\limits^{y_(max)}_{y_(min)} {m(y) \cdot g} \, dy

Where:


m(y) - Mass of the bucket as a function of height, measured in pounds.


g - Gravitational constant, measured in feet per square second. (
g = 32.174\,(ft)/(s^2))


g_(c) - lb-f to lb-m conversion factor, measured in lb-m to lb-f. (
g_(c) = 32.174\,(lbm)/(lbf))

Since leaking is constant, the mass of the bucket can be modelled by using a first-order polynomial (linear function), that is:


m(y) = m_(o)+\dot m \cdot y

Where:


m_(o) - Initial mass of the bucket, measured in pounds.


\dot m - Leaking rate, measured in pounds per feet.


y - 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:


W = (m_(o)\cdot g)/(g_(c))\int\limits^{y_(max)}_{y_(min)}\, dy + (\dot m \cdot g)/(g_(c))\int\limits^{y_(max)}_{y_(min)} {y} \, dy


W = (m_(o)\cdot g)/(g_(c))\cdot (y_(max)-y_(min)) +(\dot m \cdot g)/(2\cdot g_(c))\cdot (y_(max)-y_(min))^(2)

If
m_(o) = 15\,pd,
g = 32.174\,(ft)/(s^(2)),
g_(c) = 32.174\,(lbm)/(lbf),
\dot m = -0.3\,(pd)/(ft) and
y_(max) - y_(min) = 35\,ft, the work done lifting the bucket to the top of the building is:


W = ((15\,pd)\cdot \left(32.174\,(ft)/(s^(2)) \right))/(32.174\,(lbm)/(lbf) )\cdot (35\,ft) + (\left(-0.3\,(pd)/(ft) \right)\cdot \left(32.174\,(ft)/(s^(2)) \right))/(2\cdot \left(32.174\,(lbm)/(lbf) \right))\cdot (35\,ft)^(2)


W = 341.25\,lbf\cdot ft

The work done lifting the bucket to the top of the building is 341.25 lbf-ft.

User Eugenio Valeiras
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