Question

a heavy rope, 60 ft long, weighs 0.7 lb/ft and hangs over the edge of a building 150 ft high. (let x be the distance in feet below the top of the building. enter $x_i^*$ as $x_i$.) (a) how much work w is done in pulling the rope to the top of the building? show how to approximate the required work by a riemann sum. $lim_{n
ightarrowinfty}sum_{i = 1}^{n}(\frac{7x_i}{10})delta x$ express the work as an integral. $int_{0}^{60}(\frac{7x}{10})dx$ evaluate the integral. 1260 ft - lb (b) how much work w is done in pulling half the rope to the top of the building? how much work w is done in pulling half the rope to the top of the building? show how to approximate the required work by a riemann sum. $lim_{n
ightarrowinfty}sum_{i = 1}^{n}(\frac{7}{10}x_i)delta x$ express the work as an integral. $int_{0}^{30}(21)dx$ evaluate the integral. 945 ft - lb

Explanation:

Step1: Recall work - formula for a small segment

The work done in lifting a small segment of the rope of length $\Delta x$ at a distance $x$ from the top of the building is $dW = F\cdot dx$, where the force $F$ required to lift the segment is equal to its weight. The weight per unit length of the rope is $0.7$ lb/ft, so $F = 0.7dx$. The work done in lifting this segment to the top of the building is $dW=0.7x dx$.

Step2: For part (a) - Riemann sum and integral

To approximate the work done in pulling the entire 60 - ft rope to the top of the building using a Riemann sum, we divide the rope into $n$ sub - intervals of length $\Delta x=\frac{60}{n}$, and let $x_i$ be a sample point in the $i$th sub - interval. The Riemann sum is $\lim_{n
ightarrow\infty}\sum_{i = 1}^{n}(0.7x_i)\Delta x$. The work as an integral is $W=\int_{0}^{60}0.7x dx$.

Step3: Evaluate the integral for part (a)

Using the power rule for integration $\int x^n dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$), we have $\int_{0}^{60}0.7x dx=0.7\int_{0}^{60}x dx=0.7\times\frac{x^{2}}{2}\big|_{0}^{60}=0.35\times(60^{2}-0^{2})=1260$ ft - lb.

Step4: For part (b) - Riemann sum and integral

When pulling half of the rope (30 ft) to the top of the building, the work done in lifting a small segment of length $\Delta x$ at a distance $x$ from the top of the building is still $dW = 0.7x dx$. The Riemann sum is $\lim_{n
ightarrow\infty}\sum_{i = 1}^{n}(0.7x_i)\Delta x$ with $x_i$ in the sub - intervals of $[0,30]$. The work as an integral is $W=\int_{0}^{30}0.7x dx$.

Step5: Evaluate the integral for part (b)

Using the power rule for integration, $\int_{0}^{30}0.7x dx=0.7\times\frac{x^{2}}{2}\big|_{0}^{30}=0.35\times(30^{2}-0^{2})=315$ ft - lb.

Answer:

(a) The work done in pulling the entire rope to the top of the building:
Riemann sum: $\lim_{n
ightarrow\infty}\sum_{i = 1}^{n}(0.7x_i)\Delta x$
Integral: $\int_{0}^{60}0.7x dx$
Value: $1260$ ft - lb
(b) The work done in pulling half of the rope to the top of the building:
Riemann sum: $\lim_{n
ightarrow\infty}\sum_{i = 1}^{n}(0.7x_i)\Delta x$ (with $x_i\in[0,30]$)
Integral: $\int_{0}^{30}0.7x dx$
Value: $315$ ft - lb