# Nonconstant Acceleration Problem

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

W=1/2mV^2-1/2mV0^2
W=FΔxCos(θ)
0=N+∑Fy
Friction=μN

3. 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!

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