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A 7.06-hp motor lifts a 243-kg beam directly upward at a constant velocity from the ground to a height of 37.1 m.

How much time is required for the lift?

(Note: 1 hp = 746 W)

Answer :

The power at velocity of the engine must first be converted from horsepower to watts: Therefore, 16.69 seconds are needed for the lift.

Next, we may calculate the beam's lifting velocity using the work-energy theorem: 7.06 hp x 746 W/hp = 5271.76 W

W = ΔE = mgh

Here W is the work done by the motor, ΔE is the change in potential energy of the beam, m is the mass of the beam, g is the acceleration due to gravity, and h is the height the beam is lifted.

The velocity of the beam:

ΔE = mgh

W = ΔE/t

t = ΔE/W

v = √(2gh)

given values:

ΔE = mgh = 243 kg x 9.81 [tex]m/s^2[/tex] x 37.1 m = 88051.47 J

W = 5271.76 W

t = ΔE/W = 88051.47 J / 5271.76 W = 16.69 s

v = √(2gh) = √(2 x 9.81 m/s^2 x 37.1 m) = 26.03 m/s

Learn more about velocity visit: brainly.com/question/24445340

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