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

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Chapter1: Units, Trigonometry. And Vectors
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### 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?
   \[
   F_{\text{new}} = \_\_\_\_\_\_\_ \, \text{N}
   \]
Transcribed Image Text:### 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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