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

`-6111.11\ \text{kJ}`

`863.28\ \text{kJ}`

Step-by-step explanation:

m = Mass of automobile = 1100 kg

v = Velocity of car = 120 km/h =
`(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=(1)/(2)m(u^2-v^2)\\\Rightarrow KE=(1)/(2)* 1100* (0-((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* 9.81* 80\\\Rightarrow PE=863280\ \text{J}`

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