Question

A train with a total mass 8.00E+6kg rises 775m as it travels a distance of 55.0km along a steady slope at a constant speed of 16.0km/hr. The frictional force on the train is 0.600 percent of the weight. Find the kinetic energy of the train.

Find the total change in its potential energy.

Find the energy dissipated by kinetic friction.

Find the power output of the train engines.

Answer 1

total KE = Total Change in PE

so

KE = 0.5 mv^2

KE = 0.5* 8e6 * (16*5*5*16/(18*18))

KE = 7.9 e 7 Joules = 79 MJ

part A:

now Use Total chnage in PE = 8e6 * 9.8 * 775

PE = 6.076* 10 ^10 J

part B :Fric force = u mg

energy E = u ( mgh )

E = 0.6 * 8e6 * 9.8 * 775

E = 3.645 *10^10 J

part C: power P = Net Work Done/time

so

totol Work Done W = 6.0839 e 10 J

time elapsed is S/v = 775*18 /(16*5) = 174.35 secs

so Power =   6.0839 e 10/174.35

P =3.48 *10^8 Watts