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What is the speed of the toolbox?

  1. Feb 13, 2012 #1
    1. The problem statement, all variables and given/known data

    Starting from rest, a toolbox slides 2 m down to the edge of a roof that has a coefficient of kinetic friction of 0.25. The roof is inclined 30 degrees from the horizontal. What is the speed of the toolbox when it reaches the edge of the roof?

    2. Relevant equations

    xf=xi+vi*t+a*t^2 (I think)

    I also broke it down into components.

    Sum of forces in the x

    mgsin(theta)-fk=ma

    Sum of forces in the y

    N-mgcos(theta)=0

    3. The attempt at a solution

    I tried using this equation but i can't find the answer. I have used the position equation to find the distance but no. I want to use my components but is not giving me the mass of the object. So without the mass I am sort of lost in the problem. don;t even know where to begin! any help??
     
  2. jcsd
  3. Feb 13, 2012 #2

    PhanthomJay

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    You can use Newton 2 to find the acceleration, as you have done, but noting that fk = uN, and N=mgcos theta. The m's will cancel. Then use the kinematic equation that relates velocity with acceleration and displacement, not time.

    It may be easier to use work-energy methods, however, that is
    [itex] W_{friction} = \Delta U + \Delta K [/itex]
     
  4. Feb 13, 2012 #3
    The first step in solving any type of physics problem that the solution is not immediately obvious (and even then) is to create a force diagram.

    From the diagram, we can see that frictional force is pulling in the opposite direction of the force pulling the toolbox down the roof. (If a problem does not include a mass, assume mass=m. If the mass cancels, then it will not matter that you assumed this, if it does not, then the answer needs to include m.) Subtracting these two forces will yield the equation:

    [itex]F_{net=}mgsinθ-Nμ[/itex] which you have

    Plugging in the known variables will give [itex]F_{net}[/itex] which equals [itex]ma[/itex].

    You should now be able to see "the light at the end of the tunnel".
     
  5. Feb 13, 2012 #4
    I can't still find the answer. I still need to find N , and N just cancelled out with mgcos(theta). I divided uk/gsin(theta) to get the mass and it gets me a very small number which i doubt is the mass of a tool box , which is 0.0127 kg than the whole problem is to find the final velocity. So i took the kinematic equation and solved for t with g as my acceleration and i got 0.9 seconds than i used v=d*t and i got 3.60 m/s which is pretty close to the right answer but i'm not sure if everything i did was right. The right answer is 3.33 m/s. I'm confused .
     
  6. Feb 13, 2012 #5

    PeterO

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    Often when the mass is not given, it is because the result is independent of mass - like when you drop things and all objects begin to fall with the same acceleration.

    If you really want a mass - make one up and work your answer. Then repeat the calculation with a different mass and check for the same answer at the end.
     
  7. Feb 14, 2012 #6
    As PeterO said, the velocity of the toolbox is independant of mass in this problem. Also note that you cannot use [itex]v=d/t[/itex] here because that equation assumes constant velocity. The toolbox will be accelerating down the roof, so the velocity will be increasing.
     
  8. Feb 14, 2012 #7
    i give up i have no idea where to starts or how to do it. Is depressing. But thank for the help though...
     
  9. Feb 14, 2012 #8

    PhanthomJay

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    Don't give up so easy, you had your equation for solving acceleration correct in your very first post! Since per your last equation above, N = mgcos theta, and since fk = uN, then substitute into your first equation to get
    mgsin theta -(u)mgcos theta = ma.
    The m's cancel, g, theta, and u are known, so solve for a, the acceleration.
    Now once you know the acceleration, use the correct kinematic equation to solve for the speed as it leaves the roof.
     
  10. Feb 14, 2012 #9

    PeterO

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    Global view:

    The roof makes an angle of 30 degrees. That means for every metre of height dropped, the toolbox has move xx m along the roof [use trig to calculate xx] or if you like, for every metre the toolbox slides along the roof, it loses yy m of height [again use trig to find yy]
    [Of course you could refer to the stardard 60-30-90 triangle's sides of 2-1-√3 ; if that sentence means nothing to you, you can ignore it for now]

    In the absence of friction - the loss in Potential energy, as the toolbox slides down the roof, would appear as kinetic energy.

    With friction, some of that PE is lost - the amount given by the work done against friction.

    You need to calculate the size of the force of friction μN where N is the normal force.

    Once you know how much KE the toolbox has, you an simply find how fast it was going.

    I hope that gives you enough direction to work through the problem.
     
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