(b) If the 5.0-%u03A9 resistor represents a crane lifting a 1.0-kg mass with 25%
ID: 2112945 • Letter: #
Question
(b) If the 5.0-%u03A9 resistor represents a crane lifting a 1.0-kg mass with 25% efficiency, how long would the circuit have to run to
lift the mass to the top of the Empire State building? (c) How long would the circuit have run to lift
the 1.0-kg mass to a height twice the radius of the earth, assuming the same 25% efficiency? (Hint:
for part (b) you can use the near earth approximation to the potential energy of gravity, in part (c)
you cannot)
Explanation / Answer
B)equivalent resistance of ckt. = 4 + (2+1)||(5+1) =4 + 3*6/9 = 4+2 = 6 Ohm so total current = 12/6 = 2 A so current through 5 Ohm resistor = 2 * 3/9 = 2/3 so power dissipated through 5 ohm resistor = I^2*R = 4*5/9 = 2.22 W so work done in time t = 2.22 t work done in lifting 1 kg mass =1*g*443 = 4345.83 J but the efficiency of crane is .25 so energy from resistor utilized = 0.25*2.22t so 0.25*2.22t = 4345.83 => t = 7822.5 s = 2.173 hrs C) as height changes from R ------> 2R change in gravitational potential energy = GMm((-1/2R) - (-1/R)) = GMm(1 - (1/2))/R = GMm/2R = 31248040.137 J ; taking GM = 3.986005x10^14 m^3/s^2, R = 6,378.1 km again to calculate t we take 0.25*2.22t = 31248040.137 => t = 56247034.7187 s = 15624.176 hrs = 651 days
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