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Grade: 12th pass

                        

In factor it is desired to lift 2000 kg of metal through a distance of 12 m in 1m .find the minimum horsepower of the engine to be used.

4 years ago

Answers : (1)

Vikas TU
12273 Points
							
velocity with which it would lift => 0.5 * 2000 * v^2
and v = 0 + 10*1 = 10 m/s.
Therfore net work done would be = >  0.5 * 2000 * 10^2 + 2000* 10 * 12
                                                    => 340,000 Joule
Horse Power = > 340,000/746 = > 455.764 HP.
4 years ago
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