Question

a heavy rope, 60 ft long, weighs 0.3 lb/ft and hangs over the edge of a building 130 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→∞}∑_{i = 1}^{n}((3x_i/10))δx express the work as an integral. ∫_{0}^{60}((3x/10))dx evaluate the integral. 540 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→∞}∑_{i = 1}^{n}((3x_i/10)+9)δx express the work as an integral. ∫_{0}^{30}((3x/10)+9)dx evaluate the integral. ft - lb

Explanation:

Step1: Recall work - formula for a small segment

The force required to lift a small segment of the rope of length $\Delta x$ at a distance $x$ from the top of the building is equal to the weight of the segment. The weight - density of the rope is $\lambda=0.3$ lb/ft. So the force $F(x) = 0.3\Delta x$. The work done in lifting this small segment to the top of the building is $dW=F(x)\cdot x=0.3x\Delta x$.

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

The rope is 60 ft long. The Riemann sum for the work done in lifting the entire 60 - ft rope to the top of the building is $\lim_{n
ightarrow\infty}\sum_{i = 1}^{n}\frac{3x_i}{10}\Delta x$, where $x_i$ is a sample point in the $i$ - th sub - interval and $\frac{3x_i}{10}=0.3x_i$. The work as an integral is $W=\int_{0}^{60}0.3x dx$.

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 $\int_{0}^{60}0.3x dx=0.3\int_{0}^{60}x dx=0.3\times\frac{x^{2}}{2}\big|_{0}^{60}=0.3\times\frac{60^{2}}{2}=0.3\times\frac{3600}{2}=540$ ft - lb.

Step4: For part (b) - Analyze the situation

When pulling half of the rope (30 ft) to the top of the building, for the bottom 30 - ft of the rope, the work done in lifting a small segment of length $\Delta x$ at a distance $x$ from the top of the building:
The force required to lift a small segment of the bottom 30 - ft of the rope is still $F(x)=0.3\Delta x$. But for the top 30 - ft of the rope (which is already lifted part - way), we can consider the work done in lifting the bottom 30 - ft. The Riemann sum is $\lim_{n
ightarrow\infty}\sum_{i = 1}^{n}(\frac{3x_i}{10}+9)\Delta x$. The 9 comes from the fact that the top 30 - ft of the rope has already been lifted an average of 15 ft (and its weight is $0.3\times30 = 9$ lb). The work as an integral is $W=\int_{0}^{30}(0.3x + 9)dx$.

Step5: Evaluate the integral for part (b)

Using the sum rule of integration $\int(a + b)dx=\int a dx+\int b dx$, we have $\int_{0}^{30}(0.3x + 9)dx=\int_{0}^{30}0.3x dx+\int_{0}^{30}9dx$.
For $\int_{0}^{30}0.3x dx=0.3\times\frac{x^{2}}{2}\big|_{0}^{30}=0.3\times\frac{900}{2}=135$.
For $\int_{0}^{30}9dx=9x\big|_{0}^{30}=9\times30 = 270$.
So $W=135 + 270=405$ ft - lb.

Answer:

405 ft - lb