### Problem StatementA constant applied force \( F_p \) of 11.0 N pushes a box with a mass \( m = 6.00 \, \text{kg} \) a distance \( x = 15.0 \, \text{m} \) across a level floor. The coefficient of kinetic friction between the box and the floor is 0.110.![Diagram Explanation]- The diagram shows a box with mass \( m \) being pushed across a floor by an applied force \( F_p \). - The arrow labeled \( F_p \) indicates the direction of the applied force.- The distance \( x \) the box travels is marked by a horizontal arrow. ### Questions1. Assuming that the box starts from rest, what is the final velocity \( v_f \) of the box at the 15.0 m point? \[ v_f = \_\_\_\_\_\_\_ \, \text{m/s} \]2. If there were no friction between the box and the floor, what applied force \( F_{\text{new}} \) would give the box the same final velocity? \[ F_{\text{new}} = \_\_\_\_\_\_\_ \, \text{N} \]

Dear student We can apply work-energy theorm. Which says: Net work done = change in kinetic energy Kinetic energy is defined as : K =1/2 mv2

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A constant applied force F, of 11.0 N pushes a box with a mass m = 6.00 kg a distance x = 15.0 m across a level floor. The coefficient of kinetic friction between the box and the floor is 0.110. m Assuming that the box starts from rest, what is the final velocity vf of the box at the 15.0 m point? Uf = m/s If there were no friction between the box and the floor, what applied force Fnew Would give the box the same final velocity? Fnew N

A constant applied force F, of 11.0 N pushes a box with a mass m = 6.00 kg a distance x = 15.0 m across a level floor. The coefficient of kinetic friction between the box and the floor is 0.110. m Assuming that the box starts from rest, what is the final velocity vf of the box at the 15.0 m point? Uf = m/s If there were no friction between the box and the floor, what applied force Fnew Would give the box the same final velocity? Fnew N

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

### Problem Statement

A constant applied force \( F_p \) of 11.0 N pushes a box with a mass \( m = 6.00 \, \text{kg} \) a distance \( x = 15.0 \, \text{m} \) across a level floor. The coefficient of kinetic friction between the box and the floor is 0.110.

![Diagram Explanation]
- The diagram shows a box with mass \( m \) being pushed across a floor by an applied force \( F_p \).
- The arrow labeled \( F_p \) indicates the direction of the applied force.
- The distance \( x \) the box travels is marked by a horizontal arrow.

### Questions

1. Assuming that the box starts from rest, what is the final velocity \( v_f \) of the box at the 15.0 m point?
\[
v_f = \_\_\_\_\_\_\_ \, \text{m/s}
\]

2. If there were no friction between the box and the floor, what applied force \( F_{\text{new}} \) would give the box the same final velocity?
\[
F_{\text{new}} = \_\_\_\_\_\_\_ \, \text{N}
\]

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