what power is required to take a train weighing 200kg at a uniform speed of 60km/hr up an incline 1 in 100 if the resistance due to friction is 1/200 of the weight of the train?
2007-03-13 15:57:10 · 2 answers · asked by Joshua I 1 in Science & Mathematics ➔ Physics
The force parallel to the incline due to gravity is
Fg = 1/100 * m * g = 19.6N
The force parallel to the incline due to friction is
Ff = 1/200 * m * g = 9.8N
So total force required is
Fg + Ff = 3/100 * m * g = 29.4 N
At 60km/h the train covers 60 000 m/3600s = 16.7m/s
Work done to cover 16.7 m is
W = 29.4N * 16.7m = 491 N.m
and power required to expand this much work in one second is
P = 491N.m/1s = 491 Watt
2007-03-16 03:30:05 · answer #1 · answered by catarthur 6 · 0⤊ 0⤋
First for the incline we approximate: tg(1/100) = sin(1/100) = 1/100 and cos(1/100) = 1. Train weight force splits along the incline which is 200*sin(1/100) = 2 kg with added friction force that is force perpendicular to the incline times friction coefficient 200*cos(1/100)*1/200 = 1. Therefore total force that does the work is 3 kg. The power P is the force (that does the work)*speed. Therefore P = 3 kg * 100/6 m/sec = 50 kg*m/sec = 50*9.81 W.
2007-03-13 23:28:14 · answer #2 · answered by fernando_007 6 · 1⤊ 0⤋