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 that is 50 meters long and makes an angle of 37 degree with respect to the horizontal, as shown in the diagram below. What is the minimum amount of work that the helicopter needs to do to lift the injured person and stretcher from the bottom end of the guide rope to the top end? If the operation takes 10 seconds, what is the power delivered by the helicopter in Watts? In horsepower?

Explanation / Answer

a)

Workdone by helicopter , W = F*d*cos(theta)

= m*g*d*cos(90-37)

= 74*9.8*50*cos(53)

= 21821.8 J

b) Power of helicopter = Workdone/time taken

= 21821.8/10

= 2182.18 W

= 2182.18/(746) hp

= 74.6 hp (horse power)

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