Physics

Josh starts his sled at the top of a 3.0m high hill that has a constant slope of 25 degrees. After reaching the bottom, he slides across a horizontal patch of snow. The hill is frictionless but the coefficient of kinetic friction between his sled and the snow is .05. How far from the base of the hill does he end up? Please note that the answer in the text book is 60m, and this is the chapter involving the application of Newton's laws. We have not reached energy yet so I don't think the solution has anything involving energy in it.

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  1. dy=3.o m
    0=25'
    mk=.05
    dx=?

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  2. In order to find how far Josh ends up from the base of the hill, you have to find Josh's speed at the base of the hill. To find his speed at the base of the hill, you need to find his acceleration down the hill. The forces acting on his sled are normal and weight forces (frictionless hill). If you align your axes so that the normal force is parallel to the y-axis, we can find the x and y components of the weight force. The x component of the weight force is what is pulling the sled down the hill (wx). Acceleration of sled on the hill (ah).

    F(sled while on hill) = mah = wx = mgsin(𝚹), the masses will cancel out so, ah = gsin(𝚹) = 9.8 m/s^2(sin(25)) = 4.14 m/s^2

    Since we know the height of the hill we can find the final velocity in the y-direction (vfy), but we need to know the acceleration in the y-direction (ay).

    ay = ahsin(𝚹) = 4.14 m/s^2(sin(25)) = -1.75 m/s^2 (negative since the direction is downward)
    vfy^2 = viy^2 + 2ayΔy = (0.0 m/s)^2 + 2(-1.75 m/s^2)(-3.0 m) = 10.5 m^2/s^2
    vfy = 3.24 m/s

    Once you solve for vfy, you can find the final speed at the bottom of the hill (vs).

    vs = vfy/sin(𝚹) = 3.24 m/s / sin(25) = 7.67 m/s

    Now you need to consider the force acting on Josh and his sled on the horizontal surface. Since the sled is moving in the x direction (flat), the normal force will equal the weight force. Kinetic friction (fk)

    Fy = may = n - w --> n = w
    Fx = max = -fk = -μkn = -μkmg

    Fx = max = -μkmg (masses will cancel so)
    ax = -μkg = -0.05(9.8 m/s^2) = -0.49 m/s^2

    Now you have the acceleration in the x-direction, you can use the same kinematics equation as before to solve for Δx. vix = vs

    vfx^2 = vix^2 + 2axΔx = (0.0 m/s)^2 = (7.67 m/s)^2 + 2(-0.49 m/s^2)Δx
    0.98 m/s^2(Δx) = 58.83 m^2/s^2
    Δx = 60.0 m

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