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17 February, 21:03

At a playground, a child slides down a slide that makes a 42° angle with the horizontal direction. the coefficient of kinetic friction for the child sliding on the slide is 0.20. what is the magnitude of her acceleration during her sliding?

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  1. 17 February, 21:49
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    Assuming metric units, metre, kilogram and seconds

    Best approach: draw a free body diagram and identify forces acting on the child, which are:

    gravity, which can be decomposed into normal and parallel (to slide) components

    N=mg (cos (theta)) [pressing on slide surface]

    F=mg (sin (theta)) [pushing child downwards, also cause for acceleration]

    m=mass of child (in kg)

    g=acceleration due to gravity = 9.81 m/s^2

    theta=angle with horizontal = 42 degrees

    Similarly, kinetic friction is slowing down the child, pushing against F, and equal to

    Fr=mu*N=mu*mg (cos (theta))

    mu=coefficient of kinetic friction = 0.2

    The net force pushing child downwards along slide is therefore

    Fnet=F-Fr

    =mg (sin (theta)) - mu*mg (cos (theta))

    =mg (sin (theta) - mu*cos (theta)) [ assuming sin (theta) > mu*cos (theta) ]

    From Newton's second law,

    F=ma, or

    a=F/m

    =mg (sin (theta) - mu*cos (theta)) / m

    = g (sin (theta) - mu*cos (theta)) [ m/s^2]

    In case imperial units are used, g is approximately 32.2 feet/s^2.

    and the answer will be in the same units [ft/s^2] since sin, cos and mu are pure numbers.
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