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18 April, 10:22

the figure shows an initially stationary block of mass m on a floor. A force of magnitude 0.500mg is then applied at upward angle θ = 20∘. What is the magnitude of the acceleration of the block across the floor if the friction coefficients are (a) μs=0.610 and μk=0.500 and (b) μs=0.400 and μk=0.300

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  1. 18 April, 10:41
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    (a) 1.054 m/s²

    (b) 1.404 m/s²

    Explanation:

    0.5·m·g·cos (θ) - μs·m·g· (1 - sin (θ)) - μk·m·g· (1 - sin (θ)) = m·a

    Which gives;

    0.5·g·cos (θ) - μ·g· (1 - sin (θ) = a

    Where:

    m = Mass of the of the block

    μ = Coefficient of friction

    g = Acceleration due to gravity = 9.81 m/s²

    a = Acceleration of the block

    θ = Angle of elevation of the block = 20°

    Therefore;

    0.5*9.81·cos (20°) - μs*9.81 * (1 - sin (20°) - μk*9.81 * (1 - sin (20°) = a

    (a) When the static friction μs = 0.610 and the dynamic friction μk = 0.500, we have;

    0.5*9.81·cos (20°) - 0.610*9.81 * (1 - sin (20°) - 0.500*9.81 * (1 - sin (20°) = 1.054 m/s²

    (b) When the static friction μs = 0.400 and the dynamic friction μk = 0.300, we have;

    0.5*9.81·cos (20°) - 0.400*9.81 * (1 - sin (20°) - 0.300*9.81 * (1 - sin (20°) = 1.404 m/s².
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