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## Homework Statement

A 2 kg box is moving at 2 m/s, the box is being slowed by friciton such| Ff| = 4.3(1+x0.5) N. What is the speed of the box after moving 20 cm?

## Homework Equations

W=1/2mV^2-1/2mV0^2

W=FΔxCos(θ)

0=N+∑Fy

Friction=μN

## The Attempt at a Solution

So I realize that this is a nonconstant acceleration problem, meaning total work ends up being the integral of FΔxCos(θ), rather than just FΔx.

My knowns are:

Friction=Ff=4.2(1+x^.5)N

Vf= ?

V0= 2m/s

m=2kg

Weight=Fg=2kg*-9.8m/s^2=-19.6N

Normal=N=?

Time=?

ΔX=.2m

I can find normal using 0=N+∑Fy, where all forces pushing in the downward direction are cancelled out by the normal force. So N=-∑Fy and since there is only gravity pushing on the box, the Normal force is 19.6 N upwards.

I input the normal force into the friction formula to get:

Ff=4.2(1+x^.5)19.6=84.28+82.28x^.5

Work also equals 1/2mV^2-1/2mV0^2, inputting all I know, I get 1/2*2*V^2-1/2*2*2^2=(V^2)-4

The only force working on the system in the direction of work (x-axis) is friction, so the net force is Ff. Therefore, W=FΔxCos(θ)=FfΔxCos(180)=-FfΔx

I set this equal to (V^2)-4

-FfΔx=(V^2)-4

This is where I begin to struggle. I am pretty sure you need to integrate the left side of the equation, since this problem has nonconstant acceleration, but I'm bad at calculus and also unsure about the physics. Anyways, I ended up with

-FfΔx=-(84.28+84.28x^.5)dx, integrating from 0 to .2

56.187x^1.5+85.28x (from 0 to 0.2)=-21.8815

-21.8815=V^2-4

-17.88=V^2

.... and thats where I grind to a halt.

I'm sure I did something completely wrong, or totally interpreted the problem incorrectly but for the life of me I can't figure out what. I would really appreciate some feed back on what I did wrong and what I can do to solve this problem.

Also, this is a multiple choice problem so the choices are:

1.7 m/s

2.66 m/s

0.58 m/s

1.42 m/s

Thanks!