The coefficient of kinetic friction for a 22 kg bobsled on a track is 0.30. What force is required …?

The coefficient of kinetic friction for a 22 kg bobsled on a track is 0.30. What force is required to push it down a 6.0° incline and achieve a speed of 68 km/h at the end of 75 m?

1 Answer

  • Hello,

    Vo=0m/s , x=75m , V=18.8888889m/s , a=?

    V² = Vo² + 2ax ==> a = [V² - Vo²]/(2x) = [18.8888889² - 0²]/(2*75) = 2.3786m/s²

    From Newton's second law: F + mgsin(θ) - μmgcos(θ) = ma

    F = m[a + μgcos(θ) - gsin(θ)] = 22[2.3786 + 0.39.8cos(6°) - 9.8*sin(6°)]

    F = 94.118N

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