Problem 93 - 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 \(\mathrm{W}\) . 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 \(\mathrm{W}\) . If she expends a total of \(1.1 \times 10^{7} \mathrm{J}\) 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: Define Variables

Let \( t_w \) be the number of hours spent walking. For the rest of the day, she spends \( 24 - t_w \) hours resting.

Step 2: Set Up the Equation

The total energy expenditure can be expressed as:\[280t_w + 100(24 - t_w) = 1.1 \times 10^7\]where both terms on the left represent the energy spent walking and resting, respectively, in joules.

Step 3: Convert Hours to Seconds

Since the rates are in watts (Joules per second), we need to convert hours to seconds by multiplying by 3600 (the number of seconds in an hour). Thus, the equation becomes:\[280 \times t_w \times 3600 + 100 \times (24-t_w) \times 3600 = 1.1 \times 10^7\]

Step 4: Simplify the Equation

Simplify the equation:\[280 \times 3600 t_w + 100 \times 3600 \times 24 - 100 \times 3600 t_w = 1.1 \times 10^7\]This simplifies to:\[1008000t_w + 8640000 - 360000t_w = 1.1 \times 10^7\]

Step 5: Solve for \( t_w \)

Combine like terms:\[648000t_w + 8640000 = 1.1 \times 10^7\]Subtract 8640000 from both sides:\[648000t_w = 1.1 \times 10^7 - 8640000\]\[648000t_w = 2360000\]Divide both sides by 648000 to find \( t_w \):\[t_w = \frac{2360000}{648000}\]\[t_w \approx 3.64\]

Step 6: Convert the Answer to Hours

Since the calculation is based on hours, and we solved for \( t_w \), the time she spent walking is \( 3.64 \) hours.

Key Concepts

Physics Problem Solving

Physics problems can sometimes seem tricky, but breaking them down into manageable steps can help demystify the process. The key to solving any physics problem is understanding what you're being asked. Carefully read the question to identify what data is provided and what you need to calculate.
In our exercise, we started by defining the variables. Here, the total time is 24 hours (a full day), and we need to find out how many of those hours were spent walking. Establishing clear variables such as the time spent walking and resting allows us to set up equations that represent the real-world scenario.
Building equations from word problems requires identifying relationships between variables, often done through the lens of physics principles like conservation laws or known formulas. Once the equation is set up, the next step involves manipulating algebraic equations—combining like terms, isolating the unknown variable, and using arithmetic operations to simplify and solve the equations. This process is a central part of problem solving in physics.

Unit Conversion in Physics

Unit conversion is often necessary in physics to ensure consistency across the problem. Units need to match across all terms in an equation to perform meaningful calculations. In many physics problems, energy is given in joules, while rates of energy expenditure are provided in watts (Joules per second). This poses the challenge of converting units to ensure calculations are coherent.
Our original exercise solution involved converting hours into seconds because the power unit, watts, translates to Joules per second. Since there are 3600 seconds in an hour, multiplying the number of hours by 3600 provides the equivalent time in seconds. This conversion is crucial because it aligns with the rate of energy expenditure given in watts, allowing for accurate calculation of total energy in joules.
Always remember, consistency is key in physics. When encountering different units, identifying a common base unit—be it time, length, mass, or others—facilitates the solving process, ensuring that our results are accurate and meaningful. Unit alignment prevents errors and miscalculations that could arise from incompatible unit operations.

Work and Energy

The concepts of work and energy are central to physics and encompass a variety of everyday phenomena. Work is done when a force causes an object to move, and energy is often referred to as the capacity to do work. In physics, these concepts are often interconnected and quantified in terms of power, a measure of how quickly work is done or energy is transferred.
In our exercise, the energy expenditure rates provided in watts signify how much energy is being used per second. The problem combines these rates to determine the total work—energy expended—over the course of a day. This daily energy expenditure is calculated using the formula for work and energy, where power (in watts) is multiplied by time (in seconds) to compute energy in joules.
Understanding work and energy within the context of their mathematical relationships is vital. The rearrangement and application of the formula \[ P = \frac{W}{t} \] where \( P \) is power, \( W \) is work or energy, and \( t \) is time, illustrate these concepts in practical scenarios, showing not just theoretical understanding, but real-life applications of physics principles. This foundational knowledge underpins many branches of physics, making it essential for students to grasp thoroughly.