Question

a bucket that weighs 4 lb and a rope of negligible weight are used to draw water from a well that is 90 ft deep. the bucket is filled with 42 lb of water and is pulled up at a rate of 2.5 ft/s, but water leaks out of a hole in the bucket at a rate of 0.25 lb/s. find the work done in pulling the bucket to the top of the well. show how to approximate the required work by a riemann sum. (let x be the height in feet above the bottom of the well. enter x_i^* as x_i.) lim_(n→∞) sum_(i = 1)^n (39200x_i) δx express the work as an integral. ∫_0^90 (46) dx evaluate the integral. ft - lb

Explanation:

Step1: Determine the force function

The initial weight of the bucket with water is $4 + 42=46$ lb. The time taken to lift the bucket a distance $x$ feet is $t=\frac{x}{2.5}$ s. The amount of water that leaks out in time $t$ is $0.25t = 0.25\times\frac{x}{2.5}=0.1x$ lb. So the force $F(x)$ required to lift the bucket at height $x$ is $F(x)=(46 - 0.1x)$ lb.

Step2: Set up the work - integral formula

The work done in lifting an object through a small distance $\Delta x$ with a force $F(x)$ is $W\approx F(x)\Delta x$. The work done in lifting the bucket from $x = 0$ to $x = 90$ is given by the definite - integral $W=\int_{a}^{b}F(x)dx$. Here, $a = 0$, $b = 90$, and $F(x)=46 - 0.1x$. So the work as an integral is $\int_{0}^{90}(46 - 0.1x)dx$.

Step3: Evaluate the integral

We know that $\int_{0}^{90}(46 - 0.1x)dx=\int_{0}^{90}46dx-\int_{0}^{90}0.1xdx$.
First, $\int_{0}^{90}46dx=46x\big|_{0}^{90}=46\times90 - 46\times0 = 4140$.
Second, $\int_{0}^{90}0.1xdx=0.1\times\frac{x^{2}}{2}\big|_{0}^{90}=0.05x^{2}\big|_{0}^{90}=0.05\times(90^{2}-0^{2})=0.05\times8100 = 405$.
Then $\int_{0}^{90}(46 - 0.1x)dx=4140-405 = 3735$ ft - lb.

Answer:

3735 ft - lb