An automobile having a mass of 900 kg initially moves along a level highway at 100 km/h relative to the highway. It then climbs a hill whose crest is 50 m above the level highway and parks at a rest area located there.
a. For the automobile, determine its changes in kinetic and potential energy, each in kJ. For each quantity, kinetic energy and potential energy, specify your choice of datum and reference value at that datum. Let g= 9.81 m/s^2

ΔKE=-347.278 kJ

ΔPE= 441.45 kJ

Explanation:

given:

mass m=900 kg

the gravitational acceleration g=9.81 m/s^2

the initial velocity =100 km/h-->100*10^3/3600=27.78 m/s

height above the highway h=50 m

h1=0m

the final velocity =0 m/s

To find:

the change in kinetic energy ΔKE

the change in potential energy ΔPE

assumption:

We take the highway as a datum

solution:

ΔKE=5*m*(^2-^2)

      =-347.278 kJ

ΔPE=m*g*(h-h1)

      = 441.45 kJ

the clapeyron equation (also called the clausius-clapeyron equation) relates the slope of a reaction line on a phase diagram to fundamental thermodynamic properties.

explanation:

do you have any options?