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March 25, 2017

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A 10.3 kg block of ice slides without friction down a long track. The start of the track is 4.2 m higher than the end of the track, and the path traveled is 8.0 m. Find the speed of the block when it reaches the end of the track.

  • Physics - ,

    sinA = Y/r = 4.2 / 8 = 0.525,
    A = 31.67deg.

    Wb=10.3kg * 9.8N/kg = 100.9N @ 31.7deg = Weight of block.

    Fp = 100.9sin31.7 = 53N = Force parallel to plane.

    Ff = 0 = Force due to friction.

    Fn=Fp - Ff = 53 - 0 = 53N = Net force.

    a = Fn / m = 53 / 10.3 = 5.15m/s^2.

    Vf^2 = Vo^2 + 2ad,
    Vf^2 = 0 + 2 * 5.15 * 8 = 82.3,
    Vf = 9.1m/s.

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