Physics

A box of mass 1.50 kg is given an initial velocity of 12.5 m/s up an inclined ramp that makes an angle of 25.0o with the ground, which is horizontal. The coefficient of kinetic friction between the box and the ramp is 0.240. How long does the box slide up the ramp before it stops moving? (You do not need to worry about whether or not the box turns around and begins to slide back down the ramp after it stops, we will look at that later.)

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asked by Tyler
  1. The force supplied by kinetic friction will do work on the box, and stop the box from eventually moving any further up the ramp. Initially, the box has kinetic energy, but will eventually be stopped by the force of friction.

    Work=F*d=µk*m*g*Sin(theta)*d

    1/2mv^2=Kinetic Energy

    Since the kinetic energy will be equal to the work done by the force of kinetic energy, I can set the two equations equal to each other:

    1/2mv^2=µk*m*g*Sin(theta)*d

    Masses cancel out and the equation becomes the following:

    1/2v^2=µk*g*Sin(theta)*d

    Where

    v=12.5 m/s
    µk=0.240
    g=9.8m/s^2
    Sin(25)=0.4226
    and
    d=?

    Solve for d:

    1/2(12.5m/s)^2=(0.240)*(9.8m/s^2*(0.4226)*d

    78.125=0.994*d

    78.125/0.994=d

    d=78.6m

    This is kind of fa up the ramp, so I hope someone checks this or use this answer and setup at your own risk.

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    posted by Devron
  2. he force supplied by kinetic friction will do work on the box, and stop the box from eventually moving any further up the ramp. Initially, the box has kinetic energy, but will eventually be stopped by the force of friction.

    Work=F*d=µk*m*g*(Sin(theta)+Cos(theta)*d

    1/2mv^2=Kinetic Energy

    Since the kinetic energy will be equal to the work done by the force of kinetic energy, I can set the two equations equal to each other:

    1/2mv^2=µk*m*g*(Sin(25)+Cos(25))*d

    Masses cancel out and the equation becomes the following:

    1/2v^2=µk*g*(Sin(25)+Cos(25))*d

    Where

    v=12.5 m/s
    µk=0.240
    g=9.8m/s^2
    Sin(25)=0.4226
    Cos(25)=0.9063
    and
    d=?

    Solve for d:

    1/2(12.5m/s)^2=(0.240)*(9.8m/s^2)*(0.4226+0.9063)*d


    78.125=3.126*d

    78.125/3.126*d

    d=25.0m

    Use the following equation:

    Vf^2=Vi^2+2ad

    Where

    Vf=0m/s
    Vi=12.5m/s
    a=?
    d=25m

    Solve for a:


    0=(12.5m/s)^2+2a*(25m)

    0=156.25+50a

    -156.25/50=a

    a=-3.125m/s^2

    Solve for t:

    Vf=Vi +at

    Where

    Vf=0m/s
    Vi=12.5m/s
    a=-3.125m/s^2
    and
    t=?

    Solve for t:

    0=12.5m/s -3.125m/s^2*t

    (-12.5m/s)/-3.125m/s^2=t

    t=4s


    ******** When I looked back on another question that I answered, I saw that I messed up on a concept. When I changed/corrected myself on that post, I went back to this problem and saw that I did the same thing and that I also answered the wrong question for the problem.

    I think that I have it correct this time around.

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    posted by Devron

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