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

hi there if you are asking for conformation that wavelingth is the correct answer is wrong it is the answer is frequency. : )

i hope this you out, have a wonderful day

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.