Mathcad - 01-06

Mathcad - 01-06 - m s 2 = The gravitational constant is h...

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Thus it would require a 20 horsepower motor P in 18.182hp = P in 13.56 kW 1 hp 0.7457 kW = P in 13.558kW = P in P lift 0.30 = Since the hydraulic system is 30% efficient, the input power is: P lift 4.067 10 3 × W = P lift W pe t = t 10 sec = The lift operates in 10 seconds so the power is: W pe 4.067 10 4 × J = which yields W pe M g h = The potential energy is g 9.807
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Unformatted text preview: m s 2 = The gravitational constant is: h 1.829m = M 2.268 10 3 × kg = h 6 ft ⋅ 1 m ⋅ 3.281 ft ⋅ ⋅ = M 5000 lb ⋅ 1 kg ⋅ 2.205 lb ⋅ ⋅ = h 6 ft ⋅ = M 5000 lb ⋅ = First we need to find the potential energy obtained by lifting the car: Problem 1-6...
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This note was uploaded on 03/26/2012 for the course ECET 231 taught by Professor Piller during the Spring '12 term at Purdue.

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