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2 October, 10:07

A helicopter, starting from rest, accelerates straight up from the roof of a hospital. The lifting force does work in raising the helicopter. An 810-kg helicopter rises from rest to a speed of 7.0 m/s in a time of 3.5 s. During this time it climbs to a height of 8.2 m. What is the average power generated by the lifting force?

a) M = 810 kg

b) V0 = 0 m/s

c) Vf = 7 m/s

d) t = 3.5 s

e) y = 8.2 m

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Answers (1)
  1. 2 October, 10:40
    0
    24,267.6 watts

    Explanation:

    from the question we are given the following:

    mass (m) = 810 kg

    final velocity (v) = 7 m/s

    initial velocity (u) = 0 m/s

    time (t) = 3.5 s

    final height (h₁) = 8.2 m

    initial height (h₀) = 0 m

    acceleration due to gravity (g) = 9.8 m/s^{2}

    find the power

    power = / frac{work done}[time}

    and

    work done = change in kinetic energy (K. E) + change in potential energy (P. E)

    work done = (0.5 mv^{2} - 0.5 mu^{2}) + (mgh₁ - mgh₀)

    since u and h₀ are zero the work done now becomes

    work done = (0.5 mv^{2}) + (mgh₁)

    work done = (0.5 x 810 x 7^{2}) + (810 x 9.8 x 8.2)

    work done = 84, 936.6 joules

    recall that power = / frac{work done}[time}

    power = / frac{84,936.6}[3.5}

    power = 24,267.6 watts
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