In the 1950s, an experimental train, which had a mass of 2.50x10^4 kg, was powered across a level track by a jet engine that produced a thrust of 5.00x10^5 n for a distance of 509m.
a. find the work done on the train
b. find the change in kinetic energy.
c. find the final kinetic energy of the train if it started from rest.
d. find the final speed of the train if there had been no fricion.

a. 2.545 *10^8 J

b. 2.545 *10^8 J

c. 2.545 *10^8 J

d. 141.688 m/s

Explanation:

Data:

mass, m =  2.5*10^4 kg

Force, F = 5.00*10^5 N

distance, d = 509 m

a. Work, W = F*d = 5.00*10^5 N * 509 m = 2.545 *10^8 J

b. Change in kinetic energy, ΔKE = W = 2.545 *10^8 J

c. Change in kinetic energy = final kinetic energy - initial kinetic energy

If the train started from rest, initial kinetic energy = 0

Then, final kinetic energy, KEf = ΔKE = 2.545 *10^8 J

d. From the definition of kinetic energy we can get the final speed (vf) as follows:

KEf = (1/2)*m*vf^2

vf = √(KEf*2/m)

vf = √(2.545 *10^8*2/2.5*10^4 )

vf = 141.688 m/s

graph slopes whilst gas cools, then it is flat whilst conensation to liquid happens, the coolong liquid then graphs as a slope when it solidifies graph is flat

latent heat