On a planet where g = 10.0 m/s2 and air resistance is negligible, a sled is at rest on a rough inclined hill rising at 30°. The initial posi

Question

On a planet where g = 10.0 m/s2 and air resistance is negligible, a sled is at rest on a rough inclined hill rising at 30°. The initial position is 8.0 m up the hill along the slope from the bottom. The object is allowed to move and it stops on a rough horizontal surface, at a distance of 4.0 m from the bottom of the hill. The coefficient of kinetic friction on the hill is 0.40. What is the coefficient of kinetic friction between the horizontal surface and the sled?

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Jezebel 2 weeks 2021-08-30T18:59:59+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-08-30T19:01:20+00:00

    Answer:

    Coefficient of kinetic friction = 0.31

    Explanation:

    g = 10 m/s²

    distance covered up the hill, S = 8 m

    Horizontal distance covered, d = 4 m

    From the free body diagram attached:

    ma =  mgsin \theta - \mu_{k} N\\N = mgcos \theta\\ma = mgsin \theta - \mu_{k} mgcos \theta\\a = gsin \theta - \mu_{k} gcos \theta\\a = 10sin 30 - (0.4*10cos30)\\a = 1.4 m/s^2

    The velocity at the bottom can be calculated by:

    v^{2} = u^{2} + 2aS\\u = 0 m/s\\v^{2} = 2aS\\v^{2} = 2*1.54*8\\v^{2} = 24.64\\v = 4.96 m/s

    Work done through to frictional force = change in kinetic energy:

    f_{k} d = 0.5 mv^{2} \\\mu_{k} mgd = 0.5mv^{2} \\\mu_{k} = \frac{0.5v^{2} }{gd} \\\mu_{k} = \frac{0.5*4.96^{2} }{10*4} \\\mu_{k} = 0.31

    0
    2021-08-30T19:01:50+00:00

    Answer:

    Explanation:

    a is the acceleration

    μ is the coefficient of friction

    Acceleration of the object is given by

    a = g (\sin  \theta -\mu \cos \theta)\\\\=10( \sin 30 - 0.4 \cos 30)\\\\=10(0.5-0.3464)\\\\=1.54m/s^2

    Velocity at the bottom

    v^2=u^2+2as\\\\u=0\\\\v^2=2as\\\\=2*1.54*8\\\\=24.576\\\\v=4.96m/s

    after travelling 4m , its velocity becomes 0

    a=\frac{v^2-u^2}{2s}

    a=\frac{0-u^2}{2s}

    a=\frac{-(-4.96)^2}{2*4} \\\\=-3.075m/s^2

    Coefficient of kinetic friction

    μ = F/N

    =\frac{ma}{mg} \\\\=\frac{3.075}{10} \\\\=0.31

    Therefore, the Coefficient of kinetic friction is 0.31

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