A basketball player makes a jump shot. The 0.599 kg ball is released at a height of 2.18 m above the floor with a speed of 7.05 m/s. The bal

Question

A basketball player makes a jump shot. The 0.599 kg ball is released at a height of 2.18 m above the floor with a speed of 7.05 m/s. The ball goes through the net 3.10 m above the floor at a speed of 4.19 m/s. What is the work done on the ball by air resistance, a nonconservative force?

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Thái Dương 3 years 2021-07-31T04:27:16+00:00 2 Answers 104 views 0

Answers ( )

  1. diemkieu
    0
    2021-07-31T04:28:39+00:00
    Reply

    Answer:

    W = -4.22 J

    Explanation:

    Given

    m = 0.599 kg

    vi = 7.05 m/s

    yi = 2.18 m

    vf = 4.19 m/s

    yf = 3.10 m

    We apply the equations of the Principle of Energy Conservation and the Work-Energy Theorem

    W = Ef – Ei

    W = (Kf + Uf) – (Ki + Ui)

    W = (m/2)(vf² – vi²) + mg(yf – yi)

    W = (0.599 kg/2)((4.19 m/s)² – (7.05 m/s)²) + (0.599 kg)(9.81m/s²)(3.10 m – 2.18 m)

    W = -4.22 J

  2. Orla
    0
    2021-07-31T04:29:04+00:00
    Reply

    Answer:

    W_{drag} = 4.223\,J

    Explanation:

    The situation can be described by the Principle of Energy Conservation and the Work-Energy Theorem:

    U_{g,A}+K_{A} = U_{g,B} + K_{B} + W_{drag}

    The work done on the ball due to drag is:

    W_{drag} = (U_{g,A}-U_{g,B})+(K_{A}-K_{B})

    W_{drag} = m\cdot g\cdot (h_{A}-h_{B})+ \frac{1}{2}\cdot m \cdot (v_{A}^{2}-v_{B}^{2})

    W_{drag} = (0.599\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (2.18\,m-3.10\,m)+\frac{1}{2}\cdot (0.599\,kg)\cdot [(7.05\,\frac{m}{s} )^{2}-(4.19\,\frac{m}{s} )^{2}]

    W_{drag} = 4.223\,J

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