A loaded ore car has a mass of 955 kgand rolls on rails with negligible friction

Question

A loaded ore car has a mass of 955 kgand rolls on rails with negligible friction. It starts from restand is pulled up a mine shaft by a cable connected to a winch. Theshaft is inclined at 30.0 degreesabove the horizontal. The car accelerates uniformly to a speed of2.20 m/s in 13.0 s and then continuesat constant speed.

(a) What power must the winch motor providewhen the car is moving at constant speed?

(b) What maximum power must the winch motor provide?

(c) What total energy transfers out of the motor by work by thetime the car moves off the end of the track, which is of length1300 m?
(a) What power must the winch motor providewhen the car is moving at constant speed?

(b) What maximum power must the winch motor provide?

(c) What total energy transfers out of the motor by work by thetime the car moves off the end of the track, which is of length1300 m?

Explanation / Answer

Vv = V sin 30 = 2.2 * .5 = 1.1 m/s vertically W = m g h   work done in moving carheight h P = W / t = m g h / t = m gVv a) P = 955 * 9.8 * 1.1 = 1.03 * 10E4 J/s a = v / t   = 2.2 / 13 = .1692m/s2    the acceleration of thecar along the track s = 1/2 a t2 = 1/2 * .1692 * 169 = 14.3m    distance traveled during accelerationphase F = m a Wa = F s = 955 * .1692 * 14.3 = 2311J    work done in accelerating car (not liftingcar) Pa = Wa / t = 2311 / 13 =178  J/s   power applied to provideacceleration b) Pm = 1.03 * 10E4 + 178 = 1.05 * 10E4J/s   the maximum power applied The problem doesn't state but we are apparently to assume thatthe car moves off the track at 2.2 m/s The simplest way is E = m g h + 1/2 m v2 the total energy required E = 955 * 9.8 * 1300 * sin 30 - 1/2 * 955 * 2.22 =6.08 * 10E6 J We can check this Distance traveled at constant speed = 1300 - 14.3 = 1286m t = 1286 / 2.2 = 584.4 sec W (at constant speed) = P * t = 1.03 * 10E4 * 584.4 =6.02 * 10E6 J W ( to accelerate car) = 2311 J W (lifting car for the first 14.3 m) = 955 * 9.8 * 14.3* sin 30 = 6.79 * 10E4 WT = (6.02 + .002 + .07) * 10E6 = 6.09 * 10E6J       this looks OK withinrounding There are several ways to attack this problem, just considerthe energy required for each phase   a) P = 955 * 9.8 * 1.1 = 1.03 * 10E4 J/s a = v / t   = 2.2 / 13 = .1692m/s2    the acceleration of thecar along the track s = 1/2 a t2 = 1/2 * .1692 * 169 = 14.3m    distance traveled during accelerationphase F = m a Wa = F s = 955 * .1692 * 14.3 = 2311J    work done in accelerating car (not liftingcar) Pa = Wa / t = 2311 / 13 =178  J/s   power applied to provideacceleration b) Pm = 1.03 * 10E4 + 178 = 1.05 * 10E4J/s   the maximum power applied The problem doesn't state but we are apparently to assume thatthe car moves off the track at 2.2 m/s The simplest way is E = m g h + 1/2 m v2 the total energy required E = 955 * 9.8 * 1300 * sin 30 - 1/2 * 955 * 2.22 =6.08 * 10E6 J We can check this Distance traveled at constant speed = 1300 - 14.3 = 1286m t = 1286 / 2.2 = 584.4 sec W (at constant speed) = P * t = 1.03 * 10E4 * 584.4 =6.02 * 10E6 J W ( to accelerate car) = 2311 J W (lifting car for the first 14.3 m) = 955 * 9.8 * 14.3* sin 30 = 6.79 * 10E4 WT = (6.02 + .002 + .07) * 10E6 = 6.09 * 10E6J       this looks OK withinrounding There are several ways to attack this problem, just considerthe energy required for each phase  

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