Problem 21 - Applied Physics

Question

How much electric energy (in joules) is delivered in \(5.00 \mathrm{~min}\) to an electric motor that draws \(0.500 \mathrm{~A}\) from a \(6.00-\mathrm{V}\) battery?

Step-by-Step Solution

Answer

The electric energy delivered is 900 J.

Step 1: Understanding the Problem

We need to calculate the electric energy delivered to an electric motor. The given parameters are the current \(I\) of \(0.500\, \mathrm{A}\) and the voltage \(V\) of \(6.00\, \mathrm{V}\). The motor runs for \(5.00\, \mathrm{min}\). Electric energy can be calculated using the formula \(E = P \times t\) where \(P\) is the power and \(t\) is the time.

Step 2: Identify the Formula for Power

Electric power \(P\) is calculated by multiplying the current \(I\) by the voltage \(V\). The formula is \(P = I \times V\). Our given values are \(I = 0.500\, \mathrm{A}\) and \(V = 6.00\, \mathrm{V}\).

Step 3: Calculate Power

Substitute the given values into the power formula: \(P = 0.500\, \mathrm{A} \times 6.00\, \mathrm{V} = 3.00\, \mathrm{W}\). The power generated by the motor is \(3.00\, \mathrm{W}\).

Step 4: Convert Time from Minutes to Seconds

Before calculating energy, convert the time from minutes to seconds because the standard unit for energy calculation requires time in seconds. \(5.00\, \mathrm{min} \times 60\, \mathrm{s/min} = 300\, \mathrm{s}\).

Step 5: Calculate Electric Energy

Use the energy formula \(E = P \times t\), where \(P = 3.00\, \mathrm{W}\) and \(t = 300\, \mathrm{s}\). Substitute the values: \(E = 3.00\, \mathrm{W} \times 300\, \mathrm{s} = 900\, \mathrm{J}\). The electric energy delivered to the motor is \(900\, \mathrm{J}\).

Key Concepts

Electric Power

Electric power is the rate at which electrical energy is transferred by an electric circuit. Power is essential for understanding how efficiently a device like an electric motor operates. In this context, power is calculated using the equation:\(P = I \times V\).

• \(P\) represents power measured in watts (W).
• \(I\) is the current measured in amperes (A).
• \(V\) is the voltage measured in volts (V).

By multiplying the current by the voltage, you can find how much power is being used. It's a straightforward way to determine how much energy a component is using over time. In our example, with a current of \(0.500\, \mathrm{A}\) and voltage of \(6.00\, \mathrm{V}\), the power drawn by the motor is \(3.00\, \mathrm{W}\). This simple multiplication gives us great insight into how an electrical device functions.

Current and Voltage

Current and voltage are two fundamental concepts in physics, especially when dealing with electric power. Current is the flow of electric charge, while voltage is the potential difference that drives this flow.

• Current (I) — usually measured in amperes (A), is the amount of electric charge passing through a point in the circuit per unit time.
• Voltage (V) — measured in volts (V), is the difference in electric potential between two points. It's what pushes the current through the circuit.

Understanding both is crucial for identifying how circuits work. In our exercise, the motor uses a current of \(0.500\, \mathrm{A}\) and a voltage of \(6.00\, \mathrm{V}\) to perform work. Together, they determine the electric power available to the motor, which helps us calculate the energy involved.

Joules

The joule is the standard unit of energy in the International System of Units (SI). When we talk about electric energy, asking how much energy has been transferred often means answering in joules.

• A single joule (J) is defined as the amount of energy transferred when one watt of power is applied for one second. It can also be described as the energy expended when a current of one ampere passes through a resistance of one ohm for one second.

In solving the exercise, we've calculated that the motor receives \(900\, \mathrm{J}\) of energy over \(5.00\, \mathrm{min}\) (300 seconds). This calculation helps us to understand not only the power usage of the motor but also how effectively this power is converted into work or other forms of energy.

Physics Problem Solving

Physics problem-solving is all about using formulas and units to solve real-world problems. It involves a systematic approach to deduce unknown parameters using known values and relationships.

• Define the Problem: Understand what is being asked. Recognize all given variables.
• Choose the Correct Formula: Identify relevant equations such as \(P = I \times V\) or \(E = P \times t\).
• Convert Units: Ensure all measurements are in consistent units (e.g., converting minutes to seconds).
• Calculate: Substitute the known values into your formulas and solve.

This structured approach is critical not only for success in academic settings but vital for practical applications of physics in everyday technology. By practicing such techniques, you build the skill to tackle diverse and complex situations effectively.