A 25-meter long, 15-meter wide, and 5-meter deep rectangular water tank is full. Assume the density of water is 1000 kg/cu m. The work required to pump all the water over the top of the tank, in Newton-meters, is

A) $46,286,400$
B) $45,937,500$
C) $42,863,200$
D) $36,851,400$
E) $35,573,400$

Question

A 25-meter long, 15-meter wide, and 5-meter deep rectangular water tank is full. Assume the density of water is 1000 kg/cu m. The work required to pump all the water over the top of the tank, in Newton-meters, is

A) $46,286,400$
B) $45,937,500$
C) $42,863,200$
D) $36,851,400$
E) $35,573,400$

Quizplus · Accepted Answer

The work required to pump the water over the top of the tank can be calculated using the formula for work, $W = Fd$, where $F$ is the force applied and $d$ is the distance over which the force is applied. In this case, the force is equal to the weight of the water, which can be found by multiplying the volume of the water by the density of water and the acceleration due to gravity ($g = 9.8 m/s^2$). The volume of the tank is $25m \times 15m \times 5m = 1875 m^3$. The weight of the water is $1875 m^3 \times 1000 kg/m^3 \times 9.8 m/s^2 = 18,375,000 N$. The average distance the water needs to be lifted is half the depth of the tank, which is $2.5m$, since the water level drops as it is pumped out, but since it needs to be lifted over the top, the total distance is $5m + 2.5m = 7.5m$. Therefore, the work done is $18,375,000 N \times 7.5m = 137,812,500 N\cdot m$. However, none of the provided options match this calculation, indicating a mistake in my calculation regarding the average distance the water is lifted. The correct approach should consider the full depth since the question specifies pumping all the water over the top of the tank, making the distance $5m$, not the average distance. The correct calculation for work should directly consider the total volume of water, its weight due to gravity, and the height it needs to be lifted over the tank's top without averaging the lifting distance.Revising the calculation with the correct understanding: The total volume of water is $1875 m^3$, and the density of water is $1000 kg/m^3$, so the mass of the water is $1875 m^3 \times 1000 kg/m^3 = 1,875,000 kg$. The force due to gravity on this water is $1,875,000 kg \times 9.8 m/s^2 = 18,375,000 N$. The water needs to be lifted a distance of $5m$ (the depth of the tank) to be pumped over the top. Thus, the work done is $18,375,000 N \times 5m = 91,875,000 N\cdot m$. This calculation still does not match any of the provided options, indicating a fundamental misunderstanding in my explanation of how to calculate the work required correctly.The correct approach to find the work required to pump all the water over the top of the tank involves calculating the gravitational potential energy change for the entire volume of water being lifted to the tank's top. This involves the product of the mass of the water, gravitational acceleration, and the height it is lifted. The error in my explanation was in the misapplication of the height the water is lifted and misunderstanding the calculation process for the work required in this context. The correct calculation should directly apply the principles of energy work without incorrectly averaging the lifting distance for the water volume. Given the discrepancies in my calculations and explanations, the correct answer relies on understanding that the work done is equivalent to the gravitational potential energy change for lifting the water's entire volume to the tank's top, considering the correct height and the total mass of the water.

A 25-Meter Long, 15-Meter Wide, and 5-Meter Deep Rectangular Water 46,286,40046,286,40046,286,400

A 25-Meter Long, 15-Meter Wide, and 5-Meter Deep Rectangular Water $46,286,400$