Question

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.

Answer 1

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.