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A block of mass 12 kg starts from rest and slides a distance of 7 m down an inclined plane making an angle of 40◦ with the horizontal. The coefficient of sliding friction between block and plane is 0.4. The acceleration of gravity is 9.8 m/s2 . What is the net force on the block along the incline? Answer in units of N.
029 (part 2 of 3) 10.0 points What is the speed of the block after sliding 7 m? Answer in units of m/s.
030 (part 3 of 3) 10.0 points What would be its speed if friction were negligible? Answer in units of m/s.

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Zugey

  1. M*g = 12 * 9.8 = 117.6 N. = Wt. of block.

    Fp = 117.6*sin40 = 75.6 N. = Force parallel to incline.

    Fn = 117.6*Cos40 = 90.1 N. = Normal force.

    Fk = u*Fn = o.4 * 90.1 = 36.0 N = Force of kinetic friction.

    A. Fnet = Fp-Fk = 75.6 - 36.0 = 39.6 N. = Net force.

    B. Fnet = M*a.
    39.6 = 12*a, a = 3.3 m/s^2.

    V^2 = Vo^2 + 2a*d. = 0 + 2*3.3*7 = 46.2
    V = 6.8 m/s.

    C. Fnet = Fp-Fk = 75.6-0 = 75.6 N.

    Fnet = M*a.
    75.6 = 12a, a = 6.3 m/s^2.

    V^2 = Vo^2 + 2a*d = 0 + 2*6.3*7 = 88.2,
    V = 9.4 m/s.

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    Henry

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