A 2.0 kgkg wood block is launched up a wooden ramp that is inclined at a 25 ∘∘ angle. The block’s initial speed is 11 m/sm/s . The coefficie

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A 2.0 kgkg wood block is launched up a wooden ramp that is inclined at a 25 ∘∘ angle. The block’s initial speed is 11 m/sm/s . The coefficient of kinetic friction of wood on wood is μk=0.200μk=0.200. Part A What vertical height does the block reach above its starting point?

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Kiệt Gia 4 years 2021-08-30T15:53:15+00:00 1 Answers 10 views 0

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  1. Euphemia
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    2021-08-30T15:55:02+00:00
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    The B part of the question is missing and it says;

    What speed does it have when it slides back down to its starting point?

    Answer:

    A) vertical height the block reaches above its starting point = 3.568m

    B) Speed of block when it slides back down to its starting point = 6.323 m/s

    Explanation:

    A) Since inclined at an angle of 30°,thus;

    For kinetic friction, h = mg cosθ

    Thu kinetic friction force(Fk) is ;

    Fk = μmg cosθ

    Let’s choose the x-direction to be parallel to the ramp surface.

    Resolving all the forces on the block in the x-direction;

    ΣFx; μmg cosθ = -ma – mg sinθ

    Divide each term by m to get;

    μg cosθ = -a – g sinθ

    Making a the subject of the formula,

    a = -μg cosθ – g sinθ

    a = – g(μ cosθ + sinθ)

    So a = – 9.81[0.2(cos25°) + sin 25°]

    a = – 9.81[ 0.1813 + 0.4226]

    a = -5.924 m/s²

    Using equations of motion,

    v² = u² + 2as

    Now v = 0m/s while u= 10m/s and s=Δx since ramp is chosen to be parallel to x-direction

    Thus 0² = 10² + 2(-5.924)(Δx)

    So making (Δx) the subject ;

    11.85(Δx) = 100

    So, Δx = 100/11.85 = 8.44m

    Thus; vertical height which is the vertical component displacement = (Δx) x sinθ = 8.44Sin(25) = 8.44 x 0.4226 = 3.568m

    B) when the block is sliding down, the sign of the friction changes.

    Thus;

    ΣFx; – μmg cosθ = -ma – mg sinθ

    So, a = μg cosθ – g sinθ

    a = g(μ cosθ – sinθ)

    So a = 9.81[0.2(cos25°) – sin 25°]

    a = 9.81[ 0.1813 – 0.4226] = – 2.37 m/s²

    Again, using equations of motion,

    v² = u² + 2as

    So in thus case u =0 and since it is moving down the plane, the displacement will be negative, thus

    (Δx) = – 8.44m

    So, v² = 2(-2.37)(-8.44) = 40

    Thus, v = √40 = 6.323 m/s

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