An automobile having a mass of 1100 kg initially moves along a level highway at 120 km/h relative to the highway. It then climbs a hill whose crest is 80 m above the level highway and parks at a rest area located there. Use a reference with kinetic and potential energy each equal to zero for the stationary highway before the hill. Let g = 9.81 m/s^2.

For the automobile, determine its change in kinetic energy and its change in potential energy, both in kJ. For the automobile, determine its change in kinetic energy, in kJ.
a. -8594
b. -663.1
c. -6.63x10^5
d. 663.1

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boffeemadrid boffeemadrid · Answer 1 · 2021-02-12T06:20:14+01:00

Answer:

[tex]-6111.11\ \text{kJ}[/tex]

[tex]863.28\ \text{kJ}[/tex]

Explanation:

m = Mass of automobile = 1100 kg

v = Velocity of car = 120 km/h = [tex]\dfrac{120}{3.6}\ \text{m/s}[/tex]

h = Height of hill = 80 m

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

Change in kinetic energy

[tex]KE=\dfrac{1}{2}m(u^2-v^2)\\\Rightarrow KE=\dfrac{1}{2}\times 1100\times (0-(\dfrac{120}{3.6})^2)\\\Rightarrow KE=-611111.11\ \text{J}[/tex]

Change in kinetic energy is [tex]-6111.11\ \text{kJ}[/tex]

Change in potential energy is given by

[tex]PE=mgh\\\Rightarrow PE=1100\times 9.81\times 80\\\Rightarrow PE=863280\ \text{J}[/tex]

The change in potential energy is [tex]863.28\ \text{kJ}[/tex].