- #1

murrskeez

- 13

- 0

## Homework Statement

A box of mass m is sliding along a horizontal surface.

Part A) The box leaves position x=0 with a speed v

_{0}. The box is slowed by a constant frictional force until it comes to rest at a position x=x

_{1}. Find F

_{f}, the magnitude of the average frictional force that acts on the box. (express in terms of m, v

_{0}, and x

_{1}.

I found the correct answer to part A...(F

_{f}=(mv

_{0}

^{2})/2x

_{1})

Part B) After the box comes to rest at a position x

_{1}, a person starts pushing the box giving it a speed v

_{1}. When the box reaches position x

_{2}(where x

_{2}>x

_{1}), how much work W

_{p}has the person done on the box? Assume that the box reaches x

_{2}after a person has accelerated it from rest to speed v

_{1}. Express the work in terms of m,v

_{0},x

_{1},x

_{2}, and v

_{1}.

## Homework Equations

W=ΔKE

W=F*d

## The Attempt at a Solution

So I figured that the total work would be the work done in part A (-(mv

_{0}

^{2})/2x

_{1})*(x

_{1})

added to the work done between x

_{1}and x

_{2}:

-(mv

_{0}

^{2}(x

_{2}-x

_{1}))/2x

_{1}

however when I add these two solutions, it tells me that the answer needs to include v

_{1}. So then I thought to solve for v

_{1}by setting the work done between x

_{1}and x

_{2}equal to the change in kinetic energy between that time frame which gave me:

KE = (mv

_{1}

^{2})/2 = work between x

_{1}and x

_{2}

v

_{0}

^{2}=(x

_{1}v

_{1}

^{2})/-(x

_{2}-x

_{1})

so that allowed me to have v

_{1}in my final answer...but mastering physics is still telling me it's incorrect.

Any advice/help would be great.