A train of mass 200kN has a frictional resistance of 5N per kN. The speed of the
train, at the top of an inclined of 1 in 80 is 45 km/h. Find the speed of the train after
running down the incline for 1km.
Given:
"W=200\\:\\rm kN"
"F_f=1000\\:\\rm N"
"\\sin\\theta=1\/80"
"l=1000\\:\\rm m"
The net force
"F=W\\sin\\theta-F_f\\\\\n=200\\:000*1\/80-1000=1500\\:\\rm N"The acceleration of a train
"a=\\frac{F}{m}=\\frac{Fg}{W}=\\frac{1500*9.8}{200\\: 000}=0.0735\\:\\rm m\/s^2"The final velocity
"v=\\sqrt{2la}=\\sqrt{2*1000*0.0735}=12.1\\:\\rm m\/s"
Comments
Leave a comment