g A The box leaves position x=0 with speed v0. The box is slowed by a constant frictional force until it comes to rest at position x=x1. Find Ff, the magnitude of the average frictional force that acts on the box. (Since you don't know the coefficient of friction, don't include it in your answer.) Express the frictional force in terms of m, v0, and x1. View Available Hint(s) Ff = nothing Part B After the box comes to rest at position x1, a person starts pushing the box, giving it a speed v1.

 fr = m v₀² / 2 (x₁-x₀)

Explanation:

a) For this exercise we use Newton's second law  

X axis  

    - fr = ma  

Y Axis  

     N-W = 0  

     N=W

let's look for acceleration with expressions of kinematics  

      v² = v₀² - 2 a Δx  

at the point where stop v = 0

     a = v₀² / 2 Δx

let's replace  

     -fr = m (- v₀² / 2 (x₁-x₀))

      fr = m v₀² / 2 (x₁-x₀)

b)they ask for the same  

in this case part of rest  

v₁² = 0 + 2 a Δx  

a = v₁² / 2ΔX  

we write Newton's second law  

F - fr = m a  

fr = F - ma  

fr = F - m v₁² / 2Δx

(75°f − 32) × 5/9 = 23.889°c

answer would be letter choice (a),   a ball resting on a shelf. would be another example of potential energy   >   stored energy > not moving.

a ball spinning on a finger   >   kinetic energy > moving energy

a ball rolling on the ground   >   kinetic energy   > moving energy

a   ball flying through the air   > kinetic energy   >   moving energy

hope that                                         : )