A helicopter lifts a stretcher with a 74 kg accident victim in it out of a canyo

Question

A helicopter lifts a stretcher with a 74 kg accident victim in it out of a canyon by applying a vertical force on the stretcher. The stretcher is attached to a guide rope which is 50 meters long and makes an angle of 37° with respect to the horizontal.

(b) If the operation takes 10 seconds, what power is delivered by the helicopter in Joules/second? In Horsepower?

Attempt at solution: mgh 74kg*9.8m/s^2*50m=36,260J

36,260/10sec=3626J/per 10 sec or per sec why 10 sec?

36,260/60sec= 604.3J/per sec

3626W/750W= 4.8HP (I’m thinking too high maybe helicopter winches are that powerful OH WAIT NO Helicopters use their rotors blade motors too…that’s like 100HP+)

604.3W/750W= .8HP (possible think of 3/4HP winch common for ATV’s)

But you see here I’m learning a little from you guys :)

Explanation / Answer

This step is wrong:
mgh 74kg*9.8m/s^2*50m=36,260J

Here height is vertical height. It wont be equal to length of the rope since rope is inclined

Height will be = L* sin 37 or L* cos 53
so,
h = L* sin 37 = 50 * sin 37 = 30.09 m

work done = m*g*h
= 74*9.8*30.09
= 21821.8 J
-------------------------------------------------------
P = energy / time
= 21821.8 / 10 (10 sec because it took 10 sec to lift)
=2182.18 J/s
= 2182.18/750
= 2.9 HP
Answer: 2.9 hp

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