It takes a force of 53kN on the lead car of a 16-car passenger train with mass 9.1x 10^5 kg to pull it atconstant 45 m/s on level tracks. (a) What power must the locomotive provide to the lead car? (b) How much more power to the lead car than calculated in pan (c) Would be needed to give the train an acceleration of 1.5m/s^2, at the instant the train has a speed of 45m/s on levels tracks? (d)How much more power to the lead car than calculated in pan (e) would be needed to move the train opposite of a 1.5% grade (slope anglea= arctan 0.015) at a constant 45m/s?

Answer 1

1) 0.5mv1^2 - W = 0.5mv2^2
v1 = initial speed
v2 = final speed = 0
re-arrange
0.5mv1^2 = W

Work in this case is defined as umd
u = coeffecient of friction
m = mass
d = distance it travelled

since coeffecient of friction is a function of distance, work is changing, so you need to integrate

0.5mv^2 = m* integral from 0 to x of (u*d) + 0.6*m*(d-12.5)
(do a calculation to see if it stops before or after to include the second term for work.
mass is useless, and can be eliminated. final format looks like
0.5v1^2 = integral from 0 to x of (u*d) + 0.6*(d-12.5)
find u in terms of distance ((0.6-0.1)/12.5)*x + 0.1
b) if it stops before 12.5, you solved for distance previously, plug it into u eqn to get u.
c) .5v^2

Answer 2

What is 9 plus 10