Question comes from “APEX Calculus” by Gregory Hartman.

Question 11, Section 7.5

Problem Description:

A box weighing 2 lbs lifts 10 lbs of sand vertically by 50 ft. A crack in the box allows the sand to leak out such that 9 lbs of sand is in the box at the end of the trip. Assume the sand leaked out at a uniform rate. What is the total work done in lifting the box and sand?

Solution

The work \(W\) done on the box can be found via the following formula:

\[ \begin{equation} W = \int_a^b F(x) \ dx \end{equation} \]

where \(a\) and \(b\) are the beginning and end distances over which the force was applied. In this case, \(F(x)\) represents the weight of the box as it makes its journey, since this is the force required to lift the box against gravity. Given that sand leaves the box at a uniform rate, we know that \(F(x)\) must be a linear function, and we know from the problem that \(F(0) = 10 + 2 = 12\) and \(F(50) = 9 + 2 = 11\). Using the fact that the equation of a line is \(y = mx + b\), this allows to determine that \(F(x) = -\frac{1}{50}x + 12\). This equation is plotted below:

x <- seq(0,50, 0.1)
F_x <- (-1/50) * x + 12
plot(x, F_x)

Thus, the integral needed to be solved is as follows:

\[ W = \int_a^b F(x) \ dx = \int_0^{50} -\frac{1}{50}x + 12 dx \]

Solving the integral gives:

\[ \begin{align} W &= -\frac{x^2}{100} + 12 x |_0^{50} \\ W &= -\frac{50^2}{100} + 12(50) \\ W &= -\frac{2500}{100} + 600 \\ W &= 575 \ \text{lb-ft} \end{align} \]

If you wanted to solve in R:

f <- function(x){-(1/50)*x + 12}
W <- integrate(f, lower = 0, upper = 50)
print(W)
## 575 with absolute error < 6.4e-12