The 4400 lb load is hoisted by a motor and pulley system. The load starts at 2.2
ID: 2994654 • Letter: T
Question
The 4400 lb load is hoisted by a motor and pulley system. The load starts at 2.25 ft/s and the motor supplies a constant 1500 lb force to the cable raising the elevator 8 ft. The motor has an efficiency of 76%. Determine the final velocity of the elevator and the input power that must be applied to the motor at this instant. Neglect the mass of the pulleys and cables.
PLEASE use ft/s & ft/lb in final answers
I have worked it completely out, just trying to see where i took a wrong turn. Using the Work Energy thm would be greatly appreciated!
Explanation / Answer
using work energy theorem
work done = change in kinetic energy
mgh = final KE - initial KE
4400 (32.2)(8) = Final KE - [1/2(4400 *2.25*2.25)]
1133440= final KE - 11137.5
final KE = 1122302.5
thi will give final velocity = 22.58 ft/s
this is actually the output velocity hence from efficiency input velocity is found as
efficiency = output velocity / input velocity
0.76 = 22.58 / input velocity
input velocity = 29.71 ft/s
input power is thus force * velocity
= 1500 * 29.71
= 44565.8 lb ft/s
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