Physics

Block 1 (2.5kg) is on the surface of a frictionless inclined plane that makes an angle of 20 degrees with the horizontal. Assume g = 10 m/s2

magnitude of the acceleration of block 1 down the incline plane is 3.4 m/s2

How far does block 1 travel down the plane in 1.5 s after it is released?

asked by Janelle
  1. You already know the acceleration down the plan. They told you what it is.
    a = 3.4 m/s^2.

    The distance travelled in time t is
    S = (1/2) a t^2

    Just insert t = 1.5 s and compute your answer.

    posted by drwls

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