physics

A small block is released from rest at the top of a frictionless incline. The distance from the top of the incline to the bottom, measured along the incline, is 3.60 m. The vertical distance from the top of the incline to the bottom is 1.36 m. If
g = 9.80 m/s2,
what is the acceleration of the block as it slides down the incline?

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asked by Tara
  1. (1/2) m v^2 = m g h
    v^2 = 2 g h
    v^2 = 2 * 9.81* 1.36
    v = 5.17 m/s
    so the change in velocity from top to bottom is 5.17

    acceleration is constant so average speed = 5.17/2 = 2.58 m/s

    so it took 3.6 /2.58 = 1.39 seconds for the block to get down the slope
    the velocity changed from 0 to 5.17 m/s in 1.39 seconds
    so
    a = change in velocity/change in time
    = 5.17 m/s /1.39s = 3.71 m/s^2

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    posted by Damon

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