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4 April, 01:26

A 1.0 kg block slides along a frictionless horizontal surface with a speed of 7.0 m/s. After sliding a distance of 2.0 m, the block makes a smooth transition to a frictionless ramp inclined at an angle of 40° to the horizontal.

How far up the ramp does the block slide before coming momentarily to rest?

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Answers (1)
  1. 4 April, 01:53
    0
    3.9 m

    Explanation:

    The principle of work and energy

    ΔE = W Formula (1)

    where:

    ΔE: mechanical energy change (J)

    W : work of the non-conservative forces (J)

    ΔE = Ef - E₀

    Ef : final mechanical energy

    E₀ : initial mechanical energy

    Ef = K f + Uf

    E₀ = K₀ + U₀

    K = (1/2) mv² : Kinetic energy (J)

    U = mgh : Potential energy (J)

    m: mass (kg)

    v : speed (m/s)

    h: hight (m)

    Known data

    m = 1 kg : mass of the block

    v₀ = mg (h).0 m/s Initial speed of the block

    vf = 0 = Final speed of the block

    θ = 40° : angle θ of the ramp with respect to the horizontal direction

    μk=0 : coefficient of kinetic friction

    g = 9.8 m/s² : acceleration due to gravity

    Problem development

    W = 0, Because the friction force (non-conservative force) is zero

    Principle of work and energy to the Block:

    ΔE = W

    Ef - E₀ = 0 Equation (1)

    Ef = K f + Uf = (1/2) m (0) ² + mg (h) = mg (h) (Joules)

    E₀ = K₀ + U₀ = (1/2) m (7) ² + mg (0) = 24.5m (Joules)

    In the equation (1):

    Ef = E₀

    mg (h) = 24.5m

    We divided by m both sides of the equation

    g (h) = 24.5

    h = 24.5 / g

    h = 24.5 / 9.8

    h = 2.5 m

    We apply trigonometry at the ramp to calculate how far up the ramp (d) does the block slide before coming momentarily to rest:

    sinθ = h/d

    d = h / sinθ

    d = 2.5 m / sin40°

    d = 3.9 m
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