Physics

Can someone help me solve this one? I've tried it a few times, but I don't think I'm getting it right.

A block of mass M = 2.2 kg starts from rest at a height of 3.6 m on a fixed inclined plane. The coefficient of kinetic friction between the block and the incline is μ = 0.41. The incline plane makes an angle θ = 27.7 ° with the horizontal.

a) What is in m/s2 the acceleration of the block down the ramp?


b) What is in m/s the speed of the block at the bottom of the ramp?


The block continues to slide on the ground with the same coefficient of friction.

c) What is in m/s2 the acceleration of the block on the ground?


d) how far will the block slide on the ground until coming to rest?

asked by James
  1. M*g = 2.2 * 9.8 = 21.56 N. = Wt. of block.

    Fp = 21.56*sin27.7 = 10.02 N. = Force
    parallel to incline.

    Fn = 21.56*Cos27.7 = 19.09 N. = Normal
    force.

    Fk = u*Fn = 0.41 * 19.09 = 7.83 N. = Force of kinetic friction.

    a. a = (Fp-Fk)/M = (10.02-7.83)/2.2 = 0.995 m/s^2

    b. L = h/sin27.7 = 3.6/sin27.7 = 7.74 m.
    = Length of ramp.

    V^2 = Vo^2 + 2a*L = 0 + 1.99*7.74 = 15.4
    V = 3.92 m/s.





    posted by Henry

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