Problem 87 - University Physics with Modern Physics

Question

A physics student spends part of her day walking between classes or for recreation, during which time she expends energy at an average rate of 280 W. The remainder of the day she is sitting in class, studying, or resting; during these activities, she expends energy at an average rate of 100 W. If she expends a total of 1.1 \(\times\) 10\(^7\) J of energy in a 24-hour day, how much of the day did she spend walking?

Step-by-Step Solution

Answer

She spent approximately 3.64 hours walking.

Step 1: Introduction to the Energy Equation

The total energy expenditure can be expressed using the formula for power and energy: \( E = P \times t \). We have two activities with different power expenditure rates.

Step 2: Define Variables and Equations

Let \( t_w \) be the time spent walking and \( t_s \) be the time spent sitting. We know \( t_w + t_s = 24 \) hours. The energy expenditure equations for walking and sitting are: \( E_w = 280 \times t_w \) and \( E_s = 100 \times t_s \). The total energy is \( E_w + E_s = 1.1 \times 10^7 \) J.

Step 3: Convert Time to Seconds

Since power is given in Watts (Joules per second), convert 24 hours into seconds. \( 24 \times 3600 = 86400 \) seconds.

Step 4: Set Up Equations with Converted Variables

Substitute \( t_w \) and \( t_s \) in seconds. Use \( t_s = 86400 - t_w \) in the energy equation: \( 280 \times t_w + 100 \times (86400 - t_w) = 1.1 \times 10^7 \).

Step 5: Solve the Equation for \( t_w \)

Expand and simplify the equation: \( 280t_w + 100 \times 86400 - 100t_w = 1.1 \times 10^7 \). Simplify to \( 180t_w + 8640000 = 1.1 \times 10^7 \). Solve for \( t_w \): \( 180t_w = 1.1 \times 10^7 - 8640000 \). And \( t_w = \frac{2360000}{180} \approx 13111.11 \) seconds.

Step 6: Convert \( t_w \) Back to Hours

Convert \( t_w = 13111.11 \) seconds back to hours: \( \frac{13111.11}{3600} \approx 3.64 \) hours.

Key Concepts

energy equation

When we're solving problems related to energy and power, one fundamental equation often used is the energy equation: \( E = P \times t \). Here, \( E \) represents energy, \( P \) is power, and \( t \) is time. This simple formula helps us determine the amount of energy expended over a period given the power level.
In essence, power is the rate at which energy is used or transferred. By multiplying it with time, we can find the total energy consumption. To apply this in real-life scenarios, we typically need to know the power in Watts (Joules per second) and time in seconds. This equation is versatile and applicable to diverse problems in physics, especially those involving activities over various durations, just like a student's day spent dividing time between walking and sitting. Understanding this equation allows us to interrelate these quantities and solve for unknowns.

power and energy

Power and energy are interconnected, yet they have distinct meanings in physics. Power is defined as the rate at which work is done or energy is transferred. It is measured in Watts (W), where 1 Watt equals 1 Joule per second.
Energy, on the other hand, is the capacity to do work. It can exist in different forms such as kinetic, potential, thermal, and more. It's usually measured in Joules (J).
To understand their relationship, consider this: if you know how much power is being used (i.e., the rate of energy usage), you can calculate how much energy is consumed over time using the energy equation \( E = P \times t \). Thus, knowing the power and the duration of the activity, it becomes straightforward to find the total energy used, just as in our problem to find out how a student expended energy throughout the day.

time conversion

Time conversion becomes crucial when working with equations involving power, as power is measured per second. Often, we deal with time in hours, days, or minutes, so converting these into seconds is essential.
In our physics problem, the total daily time of 24 hours is converted to seconds: \( 24 \times 3600 = 86400 \) seconds. This conversion makes calculations consistent and enables us to use the power values in Watts accurately.
Similarly, after finding results in seconds, converting back to more familiar units such as hours can be essential for practical interpretations of the results, like determining how many hours the student spent walking. This back-and-forth conversion ensures clarity and accuracy in results.

physics problem solving

Successfully solving physics problems often involves a systematic approach:

• Firstly, understand the problem, identifying key information and what's being asked.
• Next, translate all verbal information into mathematical expressions or equations, using known formulas like the energy equation.
• Convert all units to ensure consistency. In this problem, time was converted from hours to seconds.
• Then, solve these equations for the unknowns, often using algebraic manipulation.
• Finally, interpret your answer. Convert units back if needed and check the answer for reasonableness.

Following these steps helps break down complex physics problems into manageable parts, enhancing comprehension and accuracy. Practice and familiarity with these techniques build confidence and skill in physics problem-solving.