Question

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

Answer 1

Answer:

$-6111.11\ \text{kJ}$

$863.28\ \text{kJ}$

Explanation:

m = Mass of automobile = 1100 kg

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

h = Height of hill = 80 m

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

Change in kinetic energy

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

Change in kinetic energy is $-6111.11\ \text{kJ}$

Change in potential energy is given by

$PE=mgh\\\Rightarrow PE=1100\times 9.81\times 80\\\Rightarrow PE=863280\ \text{J}$

The change in potential energy is $863.28\ \text{kJ}$.