Calculating Force Exerted By A Water Pump Physics Problem

#h1 Understanding Force Exerted by a Water Pump: A Physics Problem Solved

This article delves into a physics problem concerning the force exerted by a water pump. We will break down the problem, explore the underlying concepts, and arrive at the correct solution. This comprehensive guide aims to not only solve the specific problem but also enhance your understanding of the relationship between power, force, and velocity in physics.

The Problem Statement

The core of our discussion revolves around the following question:

A water pump with a power of $3.4 imes 10^2$ watts lifts water at the rate of $7.5 imes 10^{-2}$ meters/second from a water tank. What is the force exerted by the pump on the water?

The provided options are:

A. $1.5 imes 10^3$ newtons B. $2.2 \times 10^4$ newtons C. $4.5$ newtons D. $2.6 imes 10^3$ newtons

To solve this problem effectively, we need to understand the fundamental principles that govern the relationship between power, force, and velocity.

Key Concepts and Formulas

Before diving into the solution, let's refresh our understanding of the core concepts:

Power (P)

Power is the rate at which work is done or energy is transferred. In simpler terms, it tells us how quickly energy is being used or converted. The standard unit of power is the watt (W), which is equivalent to one joule per second (J/s).

Work (W)

Work is done when a force causes displacement. Mathematically, work is defined as the product of the force and the displacement in the direction of the force.

Force (F)

Force is an interaction that, when unopposed, will change the motion of an object. It can cause an object to accelerate, decelerate, or change direction. The unit of force is the newton (N).

Velocity (v)

Velocity is the rate of change of displacement with respect to time. It is a vector quantity, meaning it has both magnitude (speed) and direction. The unit of velocity is meters per second (m/s).

The Relationship Between Power, Force, and Velocity

The crucial formula that connects these concepts is:

$P = F imes v$

Where:

• P is the power (in watts)
• F is the force (in newtons)
• v is the velocity (in meters per second)

This equation states that the power exerted is equal to the product of the force applied and the velocity of the object.

Solving the Problem: A Step-by-Step Approach

Now that we have a solid grasp of the underlying principles, let's apply them to solve the given problem.

1. Identify the Given Information

From the problem statement, we have the following information:

• Power (P) = $3.4 imes 10^2$ watts
• Velocity (v) = $7.5 imes 10^{-2}$ meters/second

2. Identify the Unknown

We are asked to find the force (F) exerted by the pump on the water.

3. Apply the Formula

We will use the formula $P = F imes v$ to find the force. Rearranging the formula to solve for F, we get:

$F = \frac{P}{v}$

4. Substitute the Values

Now, substitute the given values into the formula:

$F = \frac{3.4 imes 10^2}{7.5 imes 10^{-2}}$

5. Calculate the Force

Performing the calculation:

$F = \frac{340}{0.075}$

$F \approx 4533.33$ newtons

6. Express in Scientific Notation

Expressing the result in scientific notation, we get:

$F \approx 4.5 imes 10^3$ newtons

7. Match with the Options

Comparing our result with the given options, we find that it does not directly match any of the options. Let's review our calculations to identify any potential errors.

Upon reviewing, we made a mistake in the scientific notation conversion. The correct scientific notation should be:

$F \approx 4.5 imes 10^3$ newtons

This still doesn't match any of the provided options. Let's re-evaluate the calculation and the provided options.

Upon closer inspection, there seems to be a discrepancy. Our calculated answer of approximately $4.5 imes 10^3$ N doesn't align with any of the given choices. Let's perform the calculation again with extra precision.

$F = \frac{3.4 imes 10^2}{7.5 imes 10^{-2}} = \frac{340}{0.075} = 4533.333...$

It appears there might be an error in the provided options. The closest option to our calculated result is:

D. $2.6 imes 10^3$ newtons. This is incorrect.

To ensure we haven't made an error, let's rethink about the magnitude of the values.

• Power = 340 Watts. This is a reasonable power for a small water pump.
• Velocity = 0.075 m/s. This is a slow but plausible speed for lifting water.
• Force = Power / Velocity. So we expect the Force to be in the thousands of Newtons.

Considering the original question and the provided options, it seems there may be a typo in either the problem statement or the options. The most likely scenario is that the correct answer was intended to be $4.5 imes 10^3$ N, but it was not included in the choices. So, if we were to choose the closest answer from the list, it would be D. $2.6 imes 10^3$ newtons, even though it is not entirely accurate.

Final Answer and Conclusion

Based on our calculations, the force exerted by the pump on the water is approximately $4.5 imes 10^3$ newtons. However, since this option is not available, the closest answer from the given choices is:

D. $2.6 imes 10^3$ newtons

It is important to note that this might be due to a potential error in the options provided in the question.

This exercise highlights the importance of understanding the relationship between power, force, and velocity. By applying the formula $P = F imes v$ and carefully performing the calculations, we can solve a variety of physics problems related to these concepts. Furthermore, it demonstrates the significance of critically evaluating the results and ensuring they align with the physical context of the problem.

Further Practice

To solidify your understanding, try solving similar problems with different values for power and velocity. You can also explore scenarios where you need to calculate the velocity given the power and force. Remember to pay close attention to the units and ensure consistency throughout your calculations.

By practicing regularly, you can build confidence in your problem-solving skills and develop a deeper understanding of the fundamental principles of physics.