A box of mass m is sliding along a horizontal surface.
Part A) The box leaves position x=0 with a speed v0. The box is slowed by a constant frictional force until it comes to rest at a position x=x1. Find Ff, the magnitude of the average frictional force that acts on the box. (express in terms of m, v0, and x1.
I found the correct answer to part A...(Ff=(mv02)/2x1)
Part B) After the box comes to rest at a position x1, a person starts pushing the box giving it a speed v1. When the box reaches position x2 (where x2>x1), how much work Wp has the person done on the box? Assume that the box reaches x2 after a person has accelerated it from rest to speed v1. Express the work in terms of m,v0,x1,x2, and v1.
The Attempt at a Solution
So I figured that the total work would be the work done in part A (-(mv02)/2x1)*(x1)
added to the work done between x1 and x2:
however when I add these two solutions, it tells me that the answer needs to include v1. So then I thought to solve for v1 by setting the work done between x1 and x2 equal to the change in kinetic energy between that time frame which gave me:
KE = (mv12)/2 = work between x1 and x2
so that allowed me to have v1 in my final answer...but mastering physics is still telling me it's incorrect.
Any advice/help would be great.