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Math Help - Need some help with this question with energy

  1. #1
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    Need some help with this question with energy

    A horizontal force of 80 N acts on a mass of 6 kg resting on a horizontal surface. The mass is initially at rest and covers a distance of 5 m in 0.92 s under the action of the force. Assuming there are no energy losses due to air resistance and therefore that the acceleration is constant:

    (a) Calculate the total energy expended in the acceleration.
    (b) Plot a graph of the kinetic energy of the mass against time.
    (c) Plot a graph of the kinetic energy of the mass against distance.
    (d) Calculate the coefficient of friction between the mass and the surface.
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: Need some help with this question with energy

    a) Remember that energy expended = work = force applied times distance.

    b) Calculate the mass's acceleration from the data that it moves 5 m in 0.92 seconds: from d = \frac 1 2  at^2, solve for a. Then calculate velocity as a function of time:  v = at. Finally KE as a function of time is KE= \frac 1 2 mv^2 = \frac 1 2 m(at)^2.

    c) for KE as a function of distance use the formula v^2 = 2ad, so KE = \frac 1 2 mv^2 = \frac 1 2 m(2ad).

    d) Use  \sum F = ma. From this find the sum of forces. There are two forces acting: the 80N push and the resisting force of friction. So force of friction = pushing force - ma.
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  3. #3
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    Re: Need some help with this question with energy

    Thanks for that

    Q1/
    a) Remember that energy expended = work = force applied times distance.
    80 X 5
    400N


    b) Calculate the mass's acceleration from the data that it moves 5 m in 0.92 seconds: from , solve for a. Then calculate velocity as a function of time: . Finally KE as a function of time is .
    A=dt2 / (1/2) => 9.2ms -2 then
    V = at => 9.2 x 0.92 = 8.64 then
    KE = mv2 => m (at)2 => x 6 x (9.2 x 0.92) 2 => 214.92J

    c) for KE as a function of distance use the formula , so .
    KE = mv2 = m (2ad) = x 6 x (2 x 9.2 x 5) = 270J


    d) Use . From this find the sum of forces. There are two forces acting: the 80N push and the resisting force of friction. So force of friction = pushing force - ma.
    F = uRn Assume g=9.81ms-2 Rn = w = mg = 6 x 9.81 = 58.86N
    P = F = 80N therefore P =uRn therefore u=P / Rn = 80 / 58.86 =1.35916 is the coefficient

    Does this look somewhere in ball park?
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  4. #4
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    Re: Need some help with this question with energy

    can anyone help me if im on right track with this
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  5. #5
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    Re: Need some help with this question with energy

    Quote Originally Posted by gordonmckee View Post
    Thanks for that

    Q1/
    a) Remember that energy expended = work = force applied times distance.
    80 X 5
    400N


    b) Calculate the mass's acceleration from the data that it moves 5 m in 0.92 seconds: from , solve for a. Then calculate velocity as a function of time: . Finally KE as a function of time is .
    A=dt2 / (1/2) => 9.2ms -2 then
    V = at => 9.2 x 0.92 = 8.64 then
    KE = mv2 => m (at)2 => x 6 x (9.2 x 0.92) 2 => 214.92J
    That's not what ebaines said and it is not even clear what question you are answering here. The question asked you to "Plot a graph of the kinetic energy of the mass against time." I can understand you not posting a graph here but you have not even calculated KE as a function of time.

    "from d= \frac{1}{2}at^2 , solve for a" That gives a= \frac{2d}{t^2}, NOT " a= at^2/(1/2)". a is NOT 9.2.

    c) for KE as a function of distance use the formula , so.
    KE = mv2 = m (2ad) = x 6 x (2 x 9.2 x 5) = 270J
    That's just bad arithmetic! x 6 x (2 x 9.2 x 5)= 30 x 9.2= 276 J.
    But, in any case, you were asked to "Plot a graph of the kinetic energy of the mass against distance." so you need KE as a function of d, not a single number:
    KE= (1/2)m(2ad)= (6)(a)d. As I said before, your value for a is incorrect.

    d) Use . From this find the sum of forces. There are two forces acting: the 80N push and the resisting force of friction. So force of friction = pushing force - ma.
    F = uRn Assume g=9.81ms-2

    Rn = w = mg = 6 x 9.81 = 58.86N
    P = F = 80N therefore P =uRn therefore u=P / Rn = 80 / 58.86 =1.35916 is the coefficient
    What happened to the "-ma"?

    Does this look somewhere in ball park?
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  6. #6
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    Re: Need some help with this question with energy

    This looks like something from teeside university
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  7. #7
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    Re: Need some help with this question with energy

    it is are you doing the same course
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