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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?

Short Answer

Expert verified
She spent approximately 3.64 hours walking.

Step by step solution

01

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.
02

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.
03

Convert Time to Seconds

Since power is given in Watts (Joules per second), convert 24 hours into seconds. \( 24 \times 3600 = 86400 \) seconds.
04

Set Up Equations with Converted Variables

Substitute \( t_w \) and \( t_s \) in seconds. Use \( t_s = 86400 - t_w \) in the energy equation: \( 280 \times t_w + 100 \times (86400 - t_w) = 1.1 \times 10^7 \).
05

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.
06

Convert \( t_w \) Back to Hours

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Using a cable with a tension of 1350 N, a tow truck pulls a car 5.00 km along a horizontal roadway. (a) How much work does the cable do on the car if it pulls horizontally? If it pulls at 35.0\(^\circ\) above the horizontal? (b) How much work does the cable do on the tow truck in both cases of part (a)? (c) How much work does gravity do on the car in part (a)?

An elevator has mass 600 kg, not including passengers. The elevator is designed to ascend, at constant speed, a vertical distance of 20.0 m (five floors) in 16.0 s, and it is driven by a motor that can provide up to 40 hp to the elevator. What is the maximum number of passengers that can ride in the elevator? Assume that an average passenger has mass 65.0 kg.

On an essentially frictionless, horizontal ice rink, a skater moving at 3.0 m/s encounters a rough patch that reduces her speed to 1.65 m/s due to a friction force that is 25% of her weight. Use the work\(-\)energy theorem to find the length of this rough patch.

You are a member of an Alpine Rescue Team. You must project a box of supplies up an incline of constant slope angle \(\alpha\) so that it reaches a stranded skier who is a vertical distance \(h\) above the bottom of the incline. The incline is slippery, but there is some friction present, with kinetic friction coefficient \(\mu_k\). Use the work\(-\) energy theorem to calculate the minimum speed you must give the box at the bottom of the incline so that it will reach the skier. Express your answer in terms of \(g\), \(h\), \(\mu_k\), and \(\alpha\).

A little red wagon with mass 7.00 kg moves in a straight line on a frictionless horizontal surface. It has an initial speed of 4.00 m/s and then is pushed 3.0 m in the direction of the initial velocity by a force with a magnitude of 10.0 N. (a) Use the work\(-\)energy theorem to calculate the wagon's final speed. (b) Calculate the acceleration produced by the force. Use this acceleration in the kinematic relationships of Chapter 2 to calculate the wagon's final speed. Compare this result to that calculated in part (a).

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