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A 2.5-kg block slides down a 17.0° inclined plane with constant acceleration. The block starts from rest at the top. At the bottom its velocity reaches 0.70 m/s. The length of the incline is 1.7 m.
What is the acceleration of the block?
What is the coefficient of friction between the plane and the block?

  • physics -

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  • physics -

    Given:
    M = 2.5kg. = Mass of block.
    A = 17o.
    Vo = 0 = Initial velocity.
    V = 0.70 m/s = Final velocity.
    d = 1.7 m. = Length of incline.

    M*g = 2.5 * 9.8 = 24.5 N. = Wt. of block.

    Fp = 24.5*sin17 = 7.16 N. = Force parallel with incline.

    Fn = 24.5*cos17 = 23.43 N. = Normal force.

    Fk = u*Fn = u*23.43 = 23.43u.

    a. V^2 = Vo^2 + 2a*d.
    0.7^2 = 0 + 2a*1.7,
    a = 0.144 m/s^2.

    b. Fp-Fk = M*a.
    7.16-23.43u = 2.5*0.144.
    u = ?

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