Problem 91 - University Physics with Modern Physics

Question

A pump is required to lift 800 kg of water (about 210 gallons) per minute from a well 14.0 \(\mathrm{m}\) deep and eject it with a speed of 18.0 \(\mathrm{m} / \mathrm{s}\) (a) How much work is done per minute in lifting the water? (b) How much work is done in giving the water the kinetic energy it has when ejected? (c) What must be the power output of the pump?

Step-by-Step Solution

Answer

(a) 109760 J; (b) 129600 J; (c) 3990 W.

Step 1: Understanding the Work Done in Lifting

The work done to lift the water is calculated using the formula for gravitational work, which is \( W = mgh \), where \( m \) is the mass, \( g = 9.8 \, \mathrm{m/s}^2 \) is the acceleration due to gravity, and \( h \) is the height. Here, \( m = 800 \, \mathrm{kg} \) and \( h = 14 \, \mathrm{m} \).

Step 2: Calculate Work Done in Lifting

Substitute the given values into the formula: \( W = 800 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s}^2 \times 14 \, \mathrm{m} = 109760 \, \mathrm{J} \). The work done per minute in lifting the water is 109760 J.

Step 3: Understanding Kinetic Work Done

The work done to give the water kinetic energy is calculated using the kinetic energy formula, \( K.E. = \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is the velocity at which the water is ejected. Here, \( m = 800 \, \mathrm{kg} \) and \( v = 18.0 \, \mathrm{m/s} \).

Step 4: Calculate Work Done for Kinetic Energy

Substitute the given values into the kinetic energy formula: \( K.E. = \frac{1}{2} \times 800 \, \mathrm{kg} \times (18.0 \, \mathrm{m/s})^2 = 129600 \, \mathrm{J} \). The work done in giving the water kinetic energy is 129600 J.

Step 5: Understanding Power Output

Power is the rate of doing work over time, calculated by the formula \( P = \frac{W}{t} \), where \( t \) is time. The total work done includes both lifting and kinetic energy work. Since we calculate per minute, \( t = 60 \, \mathrm{s} \).

Step 6: Calculate Total Power Output

The total work done in one minute is the sum of lifting and kinetic energy work: \( W = 109760 \, \mathrm{J} + 129600 \, \mathrm{J} = 239360 \, \mathrm{J} \). The power output of the pump is \( P = \frac{239360 \, \mathrm{J}}{60 \, \mathrm{s}} = 3989.33 \, \mathrm{W} \) or approximately 3990 W.

Key Concepts

Gravitational Work

Gravitational work is the energy required to move an object against the force of gravity. This is a concept often encountered when dealing with lifting objects vertically.
To calculate the work done by gravity, we use the formula:

• \( W = mgh \)

Here, \( W \) is the work done, \( m \) is the mass of the object, \( g \) is the acceleration due to gravity (standardly \( 9.8 \text{ m/s}^2 \) on Earth), and \( h \) is the height the object is lifted.
In our example, we are lifting 800 kg of water from a well that is 14 meters deep. Thus, the work done in lifting it out of the well per minute is:

• \( W = 800 \text{ kg} \times 9.8 \text{ m/s}^2 \times 14 \text{ m} = 109,760 \text{ J} \text{ (joules)} \)

This value tells us the amount of energy required each minute to overcome gravity and pump the water out of the well.

Kinetic Energy

Kinetic energy refers to the energy that an object possesses due to its motion. When an object speeds up, its kinetic energy increases.
The formula to calculate kinetic energy is:

• \( K.E. = \frac{1}{2}mv^2 \)

Where \( K.E. \) is the kinetic energy, \( m \) is the mass of the object, and \( v \) is the velocity of the object.
In this context, the water is ejected at a speed of 18.0 meters per second, meaning it has acquired kinetic energy just before it leaves the pump. Using the mass of 800 kg, the work to achieve this kinetic energy is calculated as:

• \( K.E. = \frac{1}{2} \times 800 \text{ kg} \times (18.0 \text{ m/s})^2 = 129,600 \text{ J} \)

This tells us the energy it takes per minute to give the water the speed it needs to exit the pump effectively.

Power Output

Power output is a measure of how much work is being done over a specific period of time. In physics, power is measured in watts (W), where one watt is equivalent to one joule per second.
For situations involving machines or engines, like pumps, understanding power output is crucial for ensuring that devices operate efficiently. The formula for calculating power is:

• \( P = \frac{W}{t} \)

Where \( P \) is the power, \( W \) is the work done, and \( t \) is the time over which the work is done.
In our example, the total work done by the pump per minute combines both lifting the water and giving it kinetic energy. Adding these together gives us a total work of 239,360 J. Since this is done per minute, we convert this to watts using 60 seconds (as there are 60 seconds in a minute):

• \( P = \frac{239,360 \text{ J}}{60 \text{ s}} = 3,989.33 \text{ W} \)

Thus, the pump must have a power output of approximately 3990 watts to perform these tasks continuously.