My 1954 Kg Honda Odyssey accelerates from 1.15 m/s to 12.5 m/s. A) Determine the amount of work performed on the minivan by it’s engine. B)

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My 1954 Kg Honda Odyssey accelerates from 1.15 m/s to 12.5 m/s. A) Determine the amount of work performed on the minivan by it’s engine. B) If the acceleration occurs over a distance of 0.375 km, determine the force exerted on the minivan by it’s engine. C) Calculate the net work required to bring the minivan to a stop.

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Gia Bảo 3 years 2021-08-24T03:40:50+00:00 1 Answers 2 views 0

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  1. thuhuong
    0
    2021-08-24T03:41:51+00:00
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    Answer:

    A) W = 151364.2[J]; B) F = 403.63[N]; C) W = – 152656.3[J]

    Explanation:

    Through the theorem of work and energy we can find the corresponding value of work, remembering that the sum of kinetic and potential energies at the beginning plus the work done, will be equal to the sum of kinetic and potential energies at the end.

    A)

    E_{m1}+W_{1-2}=E_{m2}\\0.5*m*v_{1}^{2}+W_{1-2}=0.5*m*v_{2}^{2}\\ 0.5*(1954)*(1.15)^{2}+W_{1-2}=0.5*(1954)*(12.5)^{2}\\ W_{1-2}=151364.2[J]

    B)

    We know that work is defined as the product of the force by the distance, in this way knowing the distance and work, we can determine the force.

    W = F * d

    151364.2 = F * (0.375*1000)

    F = 403.63[N]

    C)

    We must take the initial velocity as 12.5 m/ s, and the final velocity equals zero. Then we can calculate the work using the same work theorem and the energy of Section A.

    E_{m1}+W_{1-2}=E_{m2}\\0.5*m*v_{1}^{2}+W_{1-2}=0.5*m*v_{2}^{2}\\ 0.5*(1954)*(12.5)^{2}+W_{1-2}=0.5*(1954)*(0)^{2}\\ W_{1-2}=-152656.3[J]

    Note: The negative sign means that the work is done in a direction contrary to the movement.

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