(10%) Problem 9: An elevator motor lifts a 2750-kg load a height of 35 m in 13 s
ID: 2029792 • Letter: #
Question
(10%) Problem 9: An elevator motor lifts a 2750-kg load a height of 35 m in 13 s 50% Part (a) Find the useful power output of the elevator motor, if it also increases the speed from rest to 4.2 m/s. Assume that the total mass of the counterbalanced system is 10,000 kg - so that only 2750 kg is raised in height, but the full 10,000 kg is accelerated Grade Summary Deductions Potential 9% 91% tanO cotan0 asinacos0 atanO acotan)sinh0) tanh)cotanh(O Submissions Attempts remaining: 2 (3% per attempt) detailed view sin cosO 3% 3% 3% coshO) END DegreesRadians BACKSPACE DEL CLEAR Submit Hint Feedback I give up! Hints: 1 % deduction per hint. Hints remaining: 2 Feedback: 0% deduction per feedback 50% Part (b) How much does it cost, in dollars per hour, if the electricity cost is $0.095 per kWh?Explanation / Answer
a.
work done = Potential energy of load + change in KE for system
W = 2750 x 9.8 x 35 + (0.5) (10000) (4.2)2
W = 1031450 J
t = time taken = 13 s
P = power = W/t = 1031450/13 = 79342.3 Watt
b)
Energy consumed = Pt = 79342.3 wh = 79.342 kWh
cost per Kwh = [$] 0.095
Total Cost = 79.342 kWh x ( [$] 0.095) = 7.5
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