### Physics Problem: Friction and Force Calculation#### Problem StatementA stockroom worker pushes a box with mass 11.6 kg on a horizontal surface with a constant speed of 3.40 m/s. The coefficient of kinetic friction between the box and the surface is 0.21.---#### Part A**Question:**What horizontal force must the worker apply to maintain the motion?**Answer:**\[ F = 24 \, \text{N} \](Here, the answer has been verified as correct.)---#### Part B**Question:**If the force calculated in the previous part is removed, how far does the box slide before coming to rest?**Answer:***To be provided by the student.*The answer should be expressed with the appropriate units. The tool provides an input field for the value and units to be submitted.---### Explanation of Key Concepts#### Kinetic FrictionKinetic friction occurs when two objects are moving relative to each other and rub together. The force of kinetic friction acts opposite to the direction of motion.#### Constant SpeedWhen the box moves with a constant speed, it implies that the horizontal force applied by the worker is exactly balanced by the force of kinetic friction.#### Force CalculationThe formula for calculating the force of friction (\( F_{\text{friction}} \)) is:\[ F_{\text{friction}} = \mu_k \cdot m \cdot g \]where:- \( \mu_k \) is the coefficient of kinetic friction (0.21),- \( m \) is the mass of the box (11.6 kg),- \( g \) is the acceleration due to gravity (\( \approx 9.81 \, \text{m/s}^2 \)).The horizontal force that the worker must apply is equal to this force of friction to maintain a constant speed.#### Sliding to RestWhen the applied force is removed, the box will decelerate due to kinetic friction until it comes to rest. The distance the box slides can be found using kinematic equations, taking into account the frictional force acting on the box.---Students are encouraged to use these principles to solve Part B and deduce the sliding distance by considering the deceleration caused by the frictional force.

Given data: The mass of the box is m=11.6 kg The speed of the box is v=3.40 m/s The coefficient of…

Answered: A stockroom worker pushes a box with…

A stockroom worker pushes a box with mass 11.6 kg on a horizontal surface with a constant speed of 3.40 m/s. The coefficient of kinetic friction between the box and the surface is 0.21.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Concept explainers

Static and Kinetic Friction

Kinetic friction comes into existence when there are moving surfaces means in other words we can say that the force which comes into existing between moving surfaces is known as kinetic friction.

Question

### Physics Problem: Friction and Force Calculation

#### Problem Statement
A stockroom worker pushes a box with mass 11.6 kg on a horizontal surface with a constant speed of 3.40 m/s. The coefficient of kinetic friction between the box and the surface is 0.21.

---

#### Part A
**Question:**
What horizontal force must the worker apply to maintain the motion?

**Answer:**
\[ F = 24 \, \text{N} \]
(Here, the answer has been verified as correct.)

---

#### Part B
**Question:**
If the force calculated in the previous part is removed, how far does the box slide before coming to rest?

**Answer:**
*To be provided by the student.*

The answer should be expressed with the appropriate units. The tool provides an input field for the value and units to be submitted.

---

### Explanation of Key Concepts

#### Kinetic Friction
Kinetic friction occurs when two objects are moving relative to each other and rub together. The force of kinetic friction acts opposite to the direction of motion.

#### Constant Speed
When the box moves with a constant speed, it implies that the horizontal force applied by the worker is exactly balanced by the force of kinetic friction.

#### Force Calculation
The formula for calculating the force of friction (\( F_{\text{friction}} \)) is:
\[ F_{\text{friction}} = \mu_k \cdot m \cdot g \]
where:
- \( \mu_k \) is the coefficient of kinetic friction (0.21),
- \( m \) is the mass of the box (11.6 kg),
- \( g \) is the acceleration due to gravity (\( \approx 9.81 \, \text{m/s}^2 \)).

The horizontal force that the worker must apply is equal to this force of friction to maintain a constant speed.

#### Sliding to Rest
When the applied force is removed, the box will decelerate due to kinetic friction until it comes to rest. The distance the box slides can be found using kinematic equations, taking into account the frictional force acting on the box.

---

Students are encouraged to use these principles to solve Part B and deduce the sliding distance by considering the deceleration caused by the frictional force.

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