A concrete slab of mass 400 kg accelerates down a concrete slope inclined at 35°. The u kinetic bn the slab and slope is 0.60. Determine the acceleration of the block.

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  1. I do not care how massive the block is.
    I do care the g = about 9.81 m/s^2
    Force down slope = m g sin 35
    normal force = m g cos 35
    friction force up slope = 0.60 m g cos 35
    m g ( sin 35 - 0.60 cos 35) = m a
    a = g (sin 35 - 0.60 cos 35)

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  2. M*g = 400 * 9.8 = 3920 N. = Wt. of slab,
    Fp = 3920*sin35 = 2248.4 N. = Force parallel to incline,
    Fn = 3920*Cos35 = 3211 N. = Normal force,
    Fk = u*Fn = 0.6 * 3211 = 1927 N. = Force of kinetic friction,
    Fp-Fk = M*a,
    2248.4 - 1927 = 400*a.

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  3. ROFL :)

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  4. determine the acceleration of the block.

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  5. a=g(sin 35 -0.060 cos 35) =0.829

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