Public Transport Question (1 Viewer)

[A2RAYA]

OK THIS QUESTION APPEARED IN THE '03 PAPER AND IN MY '04 HALF YEARLY. I KNOW THE ANSWER MAY BE SIMPLE BUT IM JUST NOT SEEING IT. IF ANYONE WANTS TO POST THE ANSWER UP IT'LL PUT ME OUT OF MY MISERY AS ITS BEEN KILLING ME SINCE I FIRST

Q-13/b: A SUBURBAN TRAIN OF 400 TONNES HAS TO CLIMB A GRADIENT OF 1 IN 50 AT A CONSTANT VELOCITY OF 60KM/H

IF THE POWER REQUIRED TO OVERCOME ROLLING RESISTANCE AT THIS VELOCITY IS 450 KW, CALCULATE THE OVERALL POWER REQUIRED TO CLIMB THE GRADIENT

THERES PROBABLY A SIMPLE ANSWER TO THIS QUESTION, BUT IM JUST NOT SEEING IT

 

[Xayma]

The power required is the power to overcome resistance + the power to overcome the downwards force of gravity so you could figure it out from that.

 

[inasero]

P (to overcome friction) = 450 000W
v = 60/(3.6) m/s
m = 400 000kg
angle = inverse sin (1/50)

Train has to overcome friction and mgsin#, where # = angle of slope
mg sin# = 400 000 * 9.8 * 1/50 = 78400 N
Therefore power = P (to overcome friction) + P (to overcome weight)
= 450 000 + 78000* 60 / (3.6) (P = Fv)
= 1756666 W
= 1756 kW