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The velocity of a 1.5 kg block sliding down a frictionless inclined plane is found to be 1.13 m/s. 1.2 s later, it has a velocity of 4.37 m/s. What is the angle of the plane with respect to the horizontal (in degrees)?

  • physics -

    It is accelerating at a rate (4.37-1.13)/1.2 = 2.7 m/s^2

    The Force pushing it down the ramp is
    F = a/M = 1.8 N

    With no friction, 1.8 N must be the component of the weight down the ramp, which is M g sin A. Therefore
    1.8 N = (1.5)(9.8) sinA
    sin A = 0.1224
    A = 7.0 degrees

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