QuestionDecember 15, 2025

5. (5 marks) A circular cylinder of radius R=5 m and height H=10m half full of water.Using the density of water rho =1000kg/m^3 and earth's gravitational constant g=9.8N/kg calculate the work done in pumping all the water out the top. State your answer in scientific notation with two signficant figures, e.g. 1.4times 10^5 and with correct units. square

Solution
4.3(254 votes)

Answer

2.9 \times 10^{7} J Explanation 1. Express Work as Integral Work to lift a thin slice at depth y is dW = \text{(weight)} \times \text{(distance to top)}. Total work: W = \int_0^{H/2} \rho g A (H - y) dy, where A = \pi R^2. 2. Substitute Values A = \pi (5)^2 = 25\pi, \rho = 1000, g = 9.8, H = 10. 3. Set Up and Evaluate the Integral W = 1000 \times 9.8 \times 25\pi \int_0^5 (10 - y) dy 4. Integrate \int_0^5 (10 - y) dy = [10y - \frac{1}{2}y^2]_0^5 = 50 - 12.5 = 37.5 5. Calculate Final Value W = 1000 \times 9.8 \times 25\pi \times 37.5 \approx 1000 \times 9.8 \times 25 \times 3.1416 \times 37.5 \approx 28,879,000 J

Explanation

1. Express Work as Integral Work to lift a thin slice at depth $y$ is $dW = \text{(weight)} \times \text{(distance to top)}$. Total work: $W = \int_0^{H/2} \rho g A (H - y) dy$, where $A = \pi R^2$. 2. Substitute Values $A = \pi (5)^2 = 25\pi$, $\rho = 1000$, $g = 9.8$, $H = 10$. 3. Set Up and Evaluate the Integral $W = 1000 \times 9.8 \times 25\pi \int_0^5 (10 - y) dy$ 4. Integrate $\int_0^5 (10 - y) dy = [10y - \frac{1}{2}y^2]_0^5 = 50 - 12.5 = 37.5$ 5. Calculate Final Value $W = 1000 \times 9.8 \times 25\pi \times 37.5 \approx 1000 \times 9.8 \times 25 \times 3.1416 \times 37.5 \approx 28,879,000$ J

Click to rate:

Solution4.3(254 votes)

Answer

Explanation

Similar Questions

Solution
4.3(254 votes)