Problem 2 - Applied Physics

Question

A pump is capable of developing \(4.00 \mathrm{~kW}\) of power. How many litres of water per minute can be lifted a distance of \(35.0 \mathrm{~m} ?(1 \mathrm{~L}\) of water has a mass of \(1 \mathrm{~kg} .\) )

Step-by-Step Solution

Answer

696.8 litres of water per minute.

Step 1: Understand the Problem

We need to find out how many litres of water can be pumped per minute to a height of 35.0 m when the pump is capable of developing 4.00 kW of power. We will use the relationship between power, work, and energy to solve this problem.

Step 2: Convert Power to Suitable Units

Given power is 4.00 kW. We need to convert this into Joules per second (since 1 W = 1 J/s). So, the power is \[ 4.00 \text{ kW} = 4000 \text{ W} = 4000 \text{ J/s}. \]

Step 3: Determine the Work Done to Lift Water

The work done (in Joules) is equal to the force (weight of the water) times the height (35.0 m). The force, \( F \), is the weight, \( mg \), where \( m \) is the mass and \( g \) is the acceleration due to gravity (9.8 m/s²). Thus, the work done to lift \( m \) kilograms of water is: \[ W = m \cdot g \cdot 35.0 \text{ m}. \]

Step 4: Relate Power and Work Done

Since power is also defined as work done per unit time, the relation is: \[ P = \frac{W}{t} = 4000 \text{ J/s}. \] Given time \( t = 1 \) minute = 60 seconds, the work done in 60 seconds is: \[ W_{60s} = 4000 \times 60 = 240000 \text{ J}. \]

Step 5: Calculate Mass of Water Lifted

Use the work-energy relation \( W = m \cdot g \cdot h \) to find \( m \): \[ 240000 = m \cdot 9.8 \cdot 35.0. \] Solving for \( m \), \[ m = \frac{240000}{9.8 \times 35.0} \approx 696.8 \text{ kg}. \] Since 1 litre of water has a mass of 1 kg, this is also the number of litres.

Step 6: Conclusion

A pump with 4.00 kW of power can lift approximately 696.8 litres of water 35.0 m in one minute.

Key Concepts

Power in Physics

Power in physics is a measure of how quickly work is done or how fast energy is transferred. It gives us an understanding of how effective a machine or system is at doing work in a certain amount of time. Power is usually measured in Watts (W), where one Watt is equivalent to one Joule per second \( (1 \, ext{W} = 1 \, ext{J/s}) \).

The exercise showcases this by using a pump capable of 4.00 kW, which indicates it can do 4000 Joules of work per second. By knowing the power, we can determine how much work is done over time, which allows us to solve problems involving movements and lifting, such as lifting a mass of water over a distance. Understanding power thus plays a crucial role in tasks involving energy transfer and efficiency.

Work and Energy

Work and energy are interlinked concepts in physics. Work is done when a force moves an object through a distance, and it is defined mathematically as the product of force and distance (work \( W = F \cdot d \)).

In this problem, work is done by the pump to lift water to a height of 35.0 m. The force exerted is the weight of the water, calculated by multiplying its mass by the acceleration due to gravity \( (F = m \cdot g) \). The energy required, thus the work done (in Joules), to lift the water is given by \( W = m \cdot g \cdot h \).

This linkage allows us to determine the mass of water lifted, given the power output and the height, showcasing the relationship between work, energy, and power.

Unit Conversion

Unit conversion plays a key role in physics as it enables you to work uniformly across calculations. In the exercise, power needs conversion from kilowatts to watts (as 1 kW = 1000 W) to align with standard international units.

Additionally, time is often converted from minutes to seconds because the international system of units (SI) measures time in seconds. This ensures that calculations for power and work remain consistent. By converting 4.00 kW into 4000 W and the time from 1 minute to 60 seconds, we could compute the total work done during that minute accurately.

Accurate unit conversion is crucial, particularly in calculations involving scientific and engineering contexts, to avoid errors and ensure consistency.

Physics Calculations

Physics calculations involve applying the fundamental principles and equations of physics to solve problems. These problems usually involve multiple steps, such as first understanding the problem, gathering information, converting units, performing computations, and finally, interpreting results.

In the example provided, calculations like converting kW to W, using the work-energy principle \( W = m \cdot g \cdot h \), and solving for the mass of water lifted demonstrates the practical application of physics calculations. By breaking down the process into smaller, manageable steps, these calculations become simpler to follow.

Clear organization and careful execution in each step ensure successful problem-solving, which is essential in physics and any real-world application.