Physics

Hi I have a question in Physics

A block with mass 50kg is held on a slope. Ù friction coefficient = 0.35 between the plane and the block ..

Planets angle v with the horizontal is 25 degrees. How far will the block slide down the plane in 5 seconds after it is released (in meters)?

Thanks for the answer
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asked by Oddbjørn
  1. Var sa god
    FORCE COMPONENT DOWN SLOPE = M G SIN 25
    NORMAL FORCE ON BLOCK = M G COS 25
    FRICTION FORCE UP SLOPE = mu M G cos 25

    F = Net force down slope = M G sin 25 - mu M G cos 25

    Calculate that F

    then
    a = F/M
    d = 0 + 0 t + (1/2) a t^2

    posted by Damon

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