A 7.600 kg box is resting on a horizontal surface and attached to a 2.400 kg box by a thin, light wire that passes over a frictionless pulle

Question

A 7.600 kg box is resting on a horizontal surface and attached to a 2.400 kg box by a thin, light wire that passes over a frictionless pulley. The coefficient of kinetic friction between the box and the surface is 0.1500. The pulley has the shape of a hollow sphere of mass 1.200 kg and diameter 0.2800 m. The system is released from rest and allowed to move. The hanging box falls 0.6100 m before it hits the ground. What is the speed of the hanging block just before it hits the ground

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Hưng Khoa 4 years 2021-08-12T16:12:31+00:00 1 Answers 3 views 0

Answers ( )

  1. vankhanh
    0
    2021-08-12T16:14:00+00:00
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    Answer:

    v = 1.98 m / s

    Explanation:

    Let’s analyze this exercise a bit, initially the hanging box is at a height h, which is why it has gravitational power energy, as the system is removing the kinetic energy is zero and just when the box reaches the floor its potential energy has dropped to zero and the three bodies have kinetic energy, also between the box and the horizontal surface there is friction, so there is work. Let’s use the relationship between work and energy

    starting point. Before starting the movement

             Em₀ = U = m g h

    final point. Just before the block hit the floor

             Em_f = K = ½ M v² + ½ I w² + ½ m v²

    the speed of the two blocks must be the same to maintain the tension of the rope

    The work of the friction force

    the friction force opposes the movement so its work is negative

              W = – fr x

    the law of equilibrium is the largest block

              N-W = 0

              N = W = Mg

    we substitute

             W = – μ M g x

    the relationship between the work of the non-conservative force (friction) and the energy is

             W = Em_f – Em₀

            – μ Mg x = ½ M v² + ½ I w² + ½ m v² – mg h

    the moment of inertia of a hollow sphere is

             I = ⅔ m_s r²

    angular and linear velocity are related

             v = w r

             w = v / r

    the distance the horizontal block travels must be the same as the distance the vertical block travels

              x = h

    let’s substitute

             – μ M g h = ½ (M + m) v² + ½ (⅔ m_s r²) (v/r) ² – m g h

              (- μ M + m) g h = ½ (M + m + ⅔ m_s) v²

              v² = \frac{2(\mu \ M + m ) \ g h }{ M +m + \frac{2}{3} m_s}

    let’s calculate

              v² = 2 (-0.15 7.6 +2.4) 9.8 0.61 / (7.6 + 2.4 + 2/3 1.2)

              v = \sqrt{\frac{42.32}{ 10.8} }

              v = 1.98 m / s

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