Ahelicopter, 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

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

prolly a beacuse the momentuem the car has increases while going down hill