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Math Help - kinematics / dynamics question

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    kinematics / dynamics question

    A mass of 75 kg is on an inclined plane 30 degrees to the horizontal.
    This plane is located 2 m vertically above the ground.
    The object reaches the end of the slide at speed 6.2 m/s.
    g = 9.8 m/s^2
    How far horizontally from the end of the slide does the object land, correct to 1 tenth of a metre?
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    Quote Originally Posted by scorpion007 View Post
    A mass of 75 kg is on an inclined plane 30 degrees to the horizontal.
    This plane is located 2 m vertically above the ground.
    The object reaches the end of the slide at speed 6.2 m/s.
    g = 9.8 m/s^2
    How far horizontally from the end of the slide does the object land, correct to 1 tenth of a metre?
    Note that we don't need to know how high the plane is.

    We know the initial speed and we know the final speed. We know that the acceleration is constant. If we knew the acceleration we could find the distance along the slide using the formula:
    v^2 = v0^2 + 2a(x - x0).

    Rule of thumb: If you don't know an acceleration, do a Newton's 2nd Law problem. So, set up a Free-Body-Diagram. I've got a coordinate system where +x is down the incline and +y is directed perpendicular to and upward from the incline. The object has a weight, w, directed downward. It also has a normal force, N, acting in the +y direction. Friction is not mentioned in the problem, nor implied in any way, so assume we can ignore this.

    Newton's 2nd Law in the +x direction is:
    (Sum)Fx = w*sin(theta) = ma (Where "theta" is the angle of incline.)
    Make sure you understand why we are using sine here and not cosine!

    w = mg, so
    mg*sin(theta) = ma

    a = g*sin(theta).

    You now have the acceleration, so you can now use the v^2 formula above to solve for x.

    -Dan
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    Quote Originally Posted by scorpion007 View Post
    A mass of 75 kg is on an inclined plane 30 degrees to the horizontal.
    This plane is located 2 m vertically above the ground.
    The object reaches the end of the slide at speed 6.2 m/s.
    g = 9.8 m/s^2
    How far horizontally from the end of the slide does the object land, correct to 1 tenth of a metre?
    When the mass leaves the end of the ramp its velocity is (5.37, -3.1) in units
    of m/s. This is because the speed is 6.2 m/s at an angle 30 degrees below
    the horizontal.

    Now the vertical equation of motion of the mass after leaving the end
    of the inclined plane is:

    h'' = -9.8

    so:

    h' = -9.8 t + v_0 = -9.8 t -3.1

    so:

    h = -4.9 t^2 - 3.1 t + 2

    (as the height of the mass is 2m at t=0, the time when the mass leaves the
    incline). The mass hits the ground when h=0, so we seek the times when

    -4.9 t^2 - 3.1 t + 2 = 0.

    The roots of this quadratic are t=-1.029, and t=0.3965. The first of these
    is clearly non-physical so the root we seek is t=0.3965.

    The horizontal equation of motion (taking x=0 at t=0) is:

    x'' = 0,

    so

    x' = 5.37

    and

    x = 5.37 t

    The mass hits the ground when t=0.3965, so the mass
    has travelled a distance of:

    x = 5.37*0.3965 ~= 2.13 m,

    or rounding to the nearest 1/10th of a metre 2.1m.

    RonL
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    Forum Admin topsquark's Avatar
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    (Sigh) I apparently didn't read the whole question. I'm not having a very good day!

    -Dan
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    Quote Originally Posted by topsquark View Post
    (Sigh) I apparently didn't read the whole question. I'm not having a very good day!

    -Dan
    It happens to all of us, hopefully not with increasing frequency with age.

    RonL
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    thank you very much, but i have 1 question:
    When you integrated x' to get x, why did you omit the constant? When you did the same for int(h') you included the constant, 2.
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    Quote Originally Posted by scorpion007 View Post
    thank you very much, but i have 1 question:
    When you integrated x' to get x, why did you omit the constant? When you did the same for int(h') you included the constant, 2.
    Because I have chosen a coordinate system such that x=0 at time 0.

    Quote Originally Posted by CaptainBlack
    The horizontal equation of motion (taking x=0 at t=0) is:
    RonL
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    Quote Originally Posted by scorpion007 View Post
    thank you very much, but i have 1 question:
    When you integrated x' to get x, why did you omit the constant? When you did the same for int(h') you included the constant, 2.
    An alternate answer is that you need an anti-derivative. Though any function whose derivative is the function works. Thus, for simplicity you chose C=0 though it is not necessary.
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    Quote Originally Posted by ThePerfectHacker View Post
    An alternate answer is that you need an anti-derivative. Though any function whose derivative is the function works. Thus, for simplicity you chose C=0 though it is not necessary.
    Any choice of c is a choice of the position of the y-axis, and is equally
    valid. So we choose the one that involves the least computation for
    the problem at hand.

    RonL
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    ah i see, thank you.
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