Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use the work\(-\)energy theorem to solve each of these problems. You can use Newton's laws to check your answers. (a) A skier moving at 5.00 m/s encounters a long, rough horizontal patch of snow having a coefficient of kinetic friction of 0.220 with her skis. How far does she travel on this patch before stopping? (b) Suppose the rough patch in part (a) was only 2.90 m long. How fast would the skier be moving when she reached the end of the patch? (c) At the base of a frictionless icy hill that rises at 25.0\(^\circ\) above the horizontal, a toboggan has a speed of 12.0 m/s toward the hill. How high vertically above the base will it go before stopping?

A 6.0-kg box moving at 3.0 m/s on a horizontal, frictionless surface runs into a light spring of force constant 75 N/cm. Use the work\(-\)energy theorem to find the maximum compression of the spring.

A block of ice with mass 2.00 kg slides 1.35 m down an inclined plane that slopes downward at an angle of 36.9\(^\circ\) below the horizontal. If the block of ice starts from rest, what is its final speed? Ignore friction.

When its 75-kW (100-hp) engine is generating full power, a small single-engine airplane with mass 700 kg gains altitude at a rate of 2.5 m/s (150 m/min, or 500 ft/min). What fraction of the engine power is being used to make the airplane climb? (The remainder is used to overcome the effects of air resistance and of inefficiencies in the propeller and engine.)

A 4.00-kg block of ice is placed against a horizontal spring that has force constant \(k\) = 200 N/m and is compressed 0.025 m. The spring is released and accelerates the block along a horizontal surface. Ignore friction and the mass of the spring. (a) Calculate the work done on the block by the spring during the motion of the block from its initial position to where the spring has returned to its uncompressed length. (b) What is the speed of the block after it leaves the spring?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free