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Nonconstant Acceleration Problem

  1. Mar 2, 2014 #1
    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!
     
  2. jcsd
  3. Mar 2, 2014 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    I don't understand why you calculate any normal forces.
    The force from friction is given in the problem statement, you don't have to calculate it (especially not with a different result...). You can assume it acts directly against the direction of motion, the problem is 1-dimensional and you don't have to care about angles.

    Do you mean ##| Ff| = 4.3(1+x^{0.5}) N##?
    What is x? The length in some unit?

    Don't forget units.
    This does not work if the force is not constant over Δx.
    You need an integration, indeed.

    I guess if you fix the errors I pointed out, the rest should be fine.
     
  4. Mar 2, 2014 #3
    Hi, thanks for the help, but I don't understand what errors you pointed out, could you clarify?

    1. Yes, that is the equation that I meant. The problem never specified what x was, but I assumed it was the distance traveled at time (t).

    2. Thanks, I won't forget units

    3. I used the integral of FΔxCos(θ) from point a to b to find the work done by a nonconstant force, as opposed to just FΔx without integration, does this not work? If it doesn't what should I do instead?

    4. Thanks for the confirmation

    5. I appreciate the help, but could you clarify the errors I made?
     
  5. Mar 2, 2014 #4
    I found out what I was doing wrong, in |Ff|=4.3(1+x0.5)N N is newtons not the normal force. If you don't multiply by the normal force, you end up with 2.89=Vf^2 and that gives 1.7 as an answer, which is one of the possible selections.
    Thanks!
     
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