%3Cul%3E%0A%3Cli%3EAn%20electric%20hoist%20consumes%205kW%20of%20power%20at%20a%

Question

%3Cul%3E%0A%3Cli%3EAn%20electric%20hoist%20consumes%205kW%20of%20power%20at%20a%20constant%20rate%20to%0Alift%20the%20500kg%20mass.%20%20Assuming%20that%20the%20hoist%20is%2080%25%0Aefficient%2C%20determine%20detmerine%20(a)%20the%20maxiumum%20speed%20reached%20by%0Athe%20mass%3B%20and%20(b)%20the%20acceleration%20of%20the%20mass%20when%20the%20speed%20has%0Areached%20one-half%20of%20its%20maximum%20value.%20(Note%20not%20a%20pulley%0Aproblem)%3C%2Fli%3E%0A%3C%2Ful%3E%0A%3Cdiv%3E%0A%3Cp%3E%3Cimg%20class%3D%22user-upload%22%20src%3D%0A%22http%3A%2F%2Fmedia.cheggcdn.com%2Fmedia%252F10f%252F10f9d9d2-5968-42eb-8ad8-25b970d0a2ba%252Fphp34gODf.png%22%0Aheight%3D%221552%22%20width%3D%221468%22%20%2F%3E%3C%2Fp%3E%0A%3C%2Fdiv%3E%0A

Explanation / Answer

Power applied = 0.8 * 5kW = 4kW = 4000 J-s

Weight = 500*9.81 = 4905 N

Let F be the force applied

F * v = 4000

a) Velocity is maximum when force is minimum, F_min = 4905 N

v_max = 4000/4905 = 0.815 m/s

b) When v = v_max /2 = 0.407m/s

F = 4000/0.407 = 9810N

accln = (9810 - 4905)/ 500 = 9.81 m/s^2

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