Question

An automobile having a mass of 884 kg initially moves along a level highway at 68 km/h relative to the highway. It then climbs a hill whose crest is 69 m above the level highway and parks at a rest area located there. For the automobile, determine its changes in kinetic energy, in kJ

Answer 1

Answer:

ΔK.E. = - 142.72 kJ

Explanation:

mass = 884 kg

initial velocity = 68 km/h = 68 \times \frac {5}{18} = 18.89 m/s

final velocity = 0 m/s

height = 69 m

change in kinetic energy :

ΔK.E. = $\dfrac{1}{2}m(v_f^2-v_i^2)$

ΔK.E. =$\dfrac{1}{2}\times 884 \times (0^2-18.89^2)$

ΔK.E. =-142,716.05 J

ΔK.E. =-142.72 kJ

hence change in  kinetic energy of the automobile is  -142.72 kJ