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Homework Help: Question regarding distance/mass/velocity

  1. Jul 1, 2010 #1
    1. The problem statement, all variables and given/known data

    A crate pushed along the floor with velocity Vi slides a distance d after the pushing force is removed.

    a)If the mass of the crate is doubled but the initial velocity is not changed, what distance does the crate slide before stopping? Explain.

    b) If the initial velocity of the crate is doubled to 2vi but the mass is not changd, what distance does the crate slide before stopping? Explain.

    *I know this question was posted before but in this case, the Professor told me I can include friction, so this changes the question completely.

    2. Relevant equations

    F=max (object is moving horizontally)

    3. The attempt at a solution
    Fnetx=nx + wx + Kfx where Kfx stands for the kinetic friction, n is the normal force, and w is weight pulling down on the object.

    Fnet= -Kf (kinetic friction opposes motion, so the value is negative

    I don't think I include the pushing force because it is removed after the initial velocity.

    Kinetics Equation:
    V2=Vi2 + 2a(Xf-Xi)
    0=Vi2 + 2a(Xf-Xi)
    -Vi2=2(Kf/m)d (where d stands for Xf- Xi)
    Doing algebra: d=-mVi2/ (2*Kf) (in this equation only the Vi is squared, in the denominator, the answer is (2 times kf). I hope that is clear.

    b) -m * 2 * Vi2/2 *Kf (in this equation its 2 times the initial velocity squared divided by 2 times Kf) therefore the 2's cancel out and I get the same answer as I did for part a: -m* Vi2/Kf=d

    I hope this makes sense. Is my work and answer correct?

    Thank you kindly for taking the time to help me! :)
  2. jcsd
  3. Jul 1, 2010 #2
    someone please Help!
  4. Jul 1, 2010 #3
    Can someone please answer this?
  5. Jul 3, 2010 #4
    Hi this is my first post. I'm started studying for A lvls physics this year, so i'm not an expert in physics, but i will try my best to help you.

    a)If the mass of the crate is doubled but the initial velocity is not changed, what distance does the crate slide before stopping? Explain.

    If frictional force is taken to double as well, then the crate will still travel a distance of d, as the resistance from friction increases in the same proportion as the increase in momentum.

    (final velocity)^2= (initial velocity)^2 + 2*(acceleration)*(displacement) therefore:

    same initial velocity, same deceleration = same displacement.

    However, if you assumed that the frictional force remained the same, which is nonetheless false in real life:
    Momentum of the crate will be doubled, since twice as much energy is required to bring it to a halt. Given the same magnitude of resistance, the time taken for the mass to stop will be doubled. Thus the displacement will be 2d.

    Looking at the v-t graph will make it clearer.

    [PLAIN]http://img713.imageshack.us/img713/9237/graphs.gif [Broken]

    b) If the initial velocity of the crate is doubled to 2vi but the mass is not changd, what distance does the crate slide before stopping? Explain.

    The initial velocity is doubled, with the same rate of deceleration. Looking at the area under the V-T graph, or using the formula (final velocity)^2= (initial velocity)^2 + 2*(acceleration)*(displacement), displacement increases by 4 times, to 4d.

    Not sure if this will help, but hope it does.
    Last edited by a moderator: May 4, 2017
  6. Jul 3, 2010 #5

    First, not "nx=0, wx=0" but "nx+wx=0". That is an difference :smile:

    V2=Vi2 + 2a(Xf-Xi)(?) I'm not sure what you mean by Xf and Xi. Final and initial location? Check units. I think it is bad equation.

    Friction force is the force slowing crate. It is depend from weight and friction coefficient (you can assume that you know it, it will be canceled).

    Now you have two ways:

    calculate distance using kinematics equations for distance and acceleration in uniformly accelerated motion
    Use energy conservation law (friction force doing work on distance d). This work is equal to initial kinetic energy.

    Traditionally :smile: I suggest second way - much easier.

    Whatever you choice you have to calculate distance in three cases:

    1) m, v
    2) 2m, v
    3) m, 2v

    by dividing them you get your answers.
    Last edited: Jul 3, 2010
  7. Jul 3, 2010 #6
    in this case, since the motion is horizontal, I though the X COMPONENT OF THE normal force = 0 and the X COMPONENT of the weight is equal to zero. I tried the problem again and for the condition (2m,v) my new answer is d= (initial velocity squared) (mass)/kinetic friction. ANd for condition (m, 2v) my answer is d= (2 times initial velocity squared) (mass)/kinetic friction. I hope these are correct. Thank you kindly for your help.
  8. Jul 3, 2010 #7
    also I have not learned about the energy conservation law yet but thanks.
  9. Jul 4, 2010 #8
    Do it ASAP :smile:. It is very useful tool.

    Also you have to use dynamic/kinematic equations. There are two most important (you know both of them: s=Vi*t + 0.5*a*t^2 and a=(Vf-Vi)/t ). Look at them as a system of two equations with two unknowns. To calculate acceleration you can use also 2'nd Newton's law. Thats all :smile:

    In 99% problems with acceleration movement you don't know two values... sometimes time and initial velocity... sometimes acceleration and distance etc. Just use this two equations and calculate unknow values.

  10. Jul 4, 2010 #9
    You are absolutely right. I tought that you wrote about forces - not x component of them.

    Rest looks like good for me.

    Remember, that in (2m, v) case friction force is different that in other cases...

    Last edited: Jul 4, 2010
  11. Jul 4, 2010 #10
    Were my equations correct, for the 2m, v case and the 2v, m case? Thank you!
  12. Jul 4, 2010 #11
    Yes and no :smile:

    Yes - equations are correct.

    No - because you don't know friction coefficient. So you cannot calculate distance directly. You should divide d(2m,v) by d(m,v) and d(m,2v) by d(m,v). You get value without units (frictial coefficient will be reduced).

    Friction force equal µmg in (m,v) and (m,2v) cases and 2µmg in (2m,v) case.
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