What power will be required to move a locomotive weighing 1.00 x 10^6N up a grade that rises 1.5m for each 80.0m of track, so that it moves with a speed of 40.0km/h?
The frictional force opposing the motion is 2.00 x 10^3N
The answer is 2.31 x 10^5 but I am quite lost as to how they got it.
2007-09-10 20:37:23 · 2 answers · asked by guy 4 in Science & Mathematics ➔ Physics
In one hours time, the locomotive will travel 40000 meters and increase its' elevation by 1.5*40000/80 = 750 meters. There are actually 2 parts to this question; How much energy does it take to move the locomotive 750 meters vertically, and how much energy does it take to push anything through 40,000 meters against a resistance of 2X10^3 N. As you know (or should know) work is measured in Joules and is (in this case) the product of force times distance. So the two pieces of work done in an hour would be
W1 = 1.00X10^6N * 750 m = 7.5X10^8 Joules and
W2 = 2.00X10^3N * 40,000m = 8X10^7 Joules so the -total- energy expended was
W= 7.5X10^8 + 8X10^7 = 8.3X10^8 J
But the question asked for the amount of -power- required. Power is defined as energy expended (or work done) per second. Since there are 3600 seconds in an hour, the power (in J/s or Watts) is
8.3X10^8/3600 = 2.305X10^5 which rounds to 2.31X10^5.
HTH
Doug
2007-09-10 21:21:45 · answer #1 · answered by doug_donaghue 7 · 1⤊ 0⤋
F=m*g*sin(alpha) + frictional force(2.00*10^3)
get it?
2007-09-11 04:06:00 · answer #2 · answered by FifiLone 2 · 0⤊ 0⤋