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12 October, 13:03

Two objects are moving at equal speed along a level, frictionless surface. The second object has twice the mass of the first object. They both slide up the same frictionless incline plane. Which object rises to a greater height?

A) Object 1 rises to the greater height because it weighs less.

B) Object 2 rises to the greater height because it possesses a larger amount of kinetic energy.

C) Object 2 rises to the greater height because it contains more mass.

D) Object 1 rises to the greater height because it possesses a smaller amount of kinetic energy.

The two objects rise to the same height.

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Answers (1)
  1. 12 October, 13:24
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    E) The two objects rise to the same height : h₁=h₂ = (v²) / (2g)

    Explanation:

    With coefficient of kinetic friction, μk = 0:

    We apply the principle of energy conservation:

    E₀ = Ef Formula (1)

    K₀+U₀ = Kf + Uf Formula (2)

    K = (1/2) * m*v² Formula (3)

    U = m*g*h Formula (4)

    Where:

    E₀:

    Initial total energy (J)

    Ef: Final total energy (J)

    K₀: Initial kinetic energy (J)

    U₀: Initial potential energy (J)

    Kf: Final kinetic energy (J)

    Uf:

    Final kinetic energy (J)

    v : speed (m/s)

    m: mass (kg)

    h : hight (m)

    Known data

    v₁=v₂=v

    m₁=m

    m₂ = 2m

    μk = 0 : coefficient of kinetic friction

    Problem development

    We apply formulas (1), (2), (3) and (4) for the first object, (1):

    E₀₁ = Ef₁

    K₀₁+U₀₁ = Kf₁ + Uf₁ U₀₁=0, Kf₁ = 0, m₁=m

    (1/2) * m*v²+0 = 0+m*g*h₁ : We eliminate m, then,

    (1/2) * v²=g*h₁

    h₁ = (v²) / (2g)

    We apply formulas (1), (2), (3) and (4) for the second object, (2):

    E₀₂ = Ef₂

    K₀₂+U₀₂ = Kf₂ + Uf₂ U₀₂=0, Kf₂ = 0, m₂=2m

    (1/2) * 2m*v²+0 = 0+2m*g*h₂ : We eliminate m, then,

    (1/2) * 2*v² = 2*g*h₂ : We divide by 2 on both sides of the equation, then,

    (1/2) * v²=g*h₂

    h₂ = (v²) / (2g)
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