13.5kg crate at rest on 27degree incline.wat force of friction is required to keep crate at a constant velocity once it begins sliding down?

B) what it’s coefficient of friction if the crate is sliding down 5m/s?

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  • Anonymous
    1 month ago

    For an object sliding down an incline,

    acceleration a = g(sinΘ - µ*cosΘ)

    "constant velocity" means a = 0. So

    0 = 9.8*(sin27º - µ*cos27º)

    µ = sin27º / cos27º = tan27º = 0.510

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  • 1 month ago

    We know the component of weight in line with the incline = mg*sin27 = 13.5*9.8*sin27 = 60N

    Since there is no acceleration, Fnet = 0 meaning the weight in line with the incline is exactly countered by the friction. 

    m*g*sin27 = m*g*µ*cos27

    sin27/cos27 = tan27 = µ = 0.51 <<<<<<<

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  • Anonymous
    1 month ago

    Force of friction must equal the force of gravity acting parallel to the slope = mass * g * sin 27 deg.

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