/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 On a cold day, Ming rubs her han... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 1

On a cold day, Ming rubs her hands together to warm them up. She presses her hands together with a force of \(5.0 \mathrm{N} .\) Each time she rubs them back and forth they move a distance of \(16 \mathrm{cm}\) with a coefficient of kinetic friction of \(0.45 .\) Assuming no heat flow to the surroundings, after she has rubbed her hands back and forth eight times, by how much has the internal energy of her hands increased?

Short Answer

Expert verified
total work done by the friction force. This is because the work done by friction turns into heat energy, which increases the internal energy of her hands. Thus: 螖E_internal = W_total = 2.88 J Therefore, the increase in internal energy of Ming's hands is 2.88 J.

Step by step solution

01

Calculate the friction force

First, we should calculate the friction force between Ming's hands. The friction force can be calculated using the equation \(F_{friction} = 渭 * F_n\), where \(F_{friction}\) is the friction force, \(渭\) is the coefficient of kinetic friction, and \(F_n\) is the normal force. In this case, the normal force is the force Ming applies on her hands, which is 5.0 N, and the coefficient of kinetic friction is 0.45. Thus: \(F_{friction} = 渭 * F_n = 0.45 * 5.0 \mathrm{N} = 2.25 \mathrm{N}\)
02

Calculate the work done by friction force

Now that we have the friction force, we can calculate the work done by this force each time Ming rubs her hands back and forth. The work done by a force acting over a distance can be calculated using the equation \(W = F * d\), where \(W\) is the work done, \(F\) is the force, and \(d\) is the distance. In this case, the force is the friction force, which is 2.25 N, and the distance is 16 cm (0.16 m). Thus: \(W = F * d = 2.25\mathrm{N} * 0.16\mathrm{m} = 0.36\mathrm{J}\)
03

Calculate the total work done

To find the total work done after Ming rubs her hands back and forth eight times, we just need to multiply the work done each time (0.36 J) by the number of times she does it (8). \(W_{total} = W * n = 0.36\mathrm{J} * 8 = 2.88\mathrm{J}\)
04

Determine the increase in internal energy

Assuming no heat flow to the surroundings, the increase in internal energy of Ming's hands is equal to the

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 91影视!

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

In a certain steam engine, the boiler temperature is \(127^{\circ} \mathrm{C}\) and the cold reservoir temperature is \(27^{\circ} \mathrm{C} .\) While this engine does \(8.34 \mathrm{kJ}\) of work, what minimum amount of heat must be discharged into the cold reservoir?
A system takes in \(550 \mathrm{J}\) of heat while performing \(840 \mathrm{J}\) of work. What is the change in internal energy of the system?
A \(0.50-\mathrm{kg}\) block of iron $[c=0.44 \mathrm{kJ} /(\mathrm{kg} \cdot \mathrm{K})]\( at \)20.0^{\circ} \mathrm{C}$ is in contact with a \(0.50-\mathrm{kg}\) block of aluminum \([c=\) $0.900 \mathrm{kJ} /(\mathrm{kg} \cdot \mathrm{K})]$ at a temperature of \(20.0^{\circ} \mathrm{C} .\) The system is completely isolated from the rest of the universe. Suppose heat flows from the iron into the aluminum until the temperature of the aluminum is \(22.0^{\circ} \mathrm{C}\) (a) From the first law, calculate the final temperature of the iron. (b) Estimate the entropy change of the system. (c) Explain how the result of part (b) shows that this process is impossible. [Hint: since the system is isolated, $\left.\Delta S_{\text {System }}=\Delta S_{\text {Universe }} .\right]$

Show that the coefficient of performance for a reversible refrigerator is \(1 /\left[\left(T_{\mathrm{H}} / T_{\mathrm{C}}\right)-1\right]\).
An inventor proposes a heat engine to propel a ship, using the temperature difference between the water at the surface and the water \(10 \mathrm{m}\) below the surface as the two reservoirs. If these temperatures are \(15.0^{\circ} \mathrm{C}\) and \(10.0^{\circ} \mathrm{C},\) respectively, what is the maximum possible efficiency of the engine?
See all solutions

Recommended explanations on Physics Textbooks

Fundamentals of Physics

Read Explanation

Atoms and Radioactivity

Read Explanation

Thermodynamics

Read Explanation

Rotational Dynamics

Read Explanation

Particle Model of Matter

Read Explanation

Kinematics Physics

Read Explanation
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.