Question

An automobile having a mass of 1100 kg initially moves along a level highway at 110 km/h relative to the highway. It then climbs a hill whose crest is 40 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/s2.
For the automobile, determine its change in kinetic energy and its change in potential energy, both in kJ.

Answer 1

Answer:

Change in kinetic energy=-513.652 KJ

Change in potential energy=431.64KJ

Explanation:

We are given that

Mass of an automobile , m=1100 kg

Initial speed, u=110 km/h=$110\times \frac{5}{18}=30.56 m/s$

Where $1 km/h=5/18 m/s$

Height , $h_2=40 m$

$g=9.81 m/s^2$

Final speed, v=0

Change in kinetic energy,$\Delta K.E=\frac{1}{2}m(v^2-u^2)$

$\Delta K.E=\frac{1}{2}(1100)(0-(30.56)^2)=-513652.48 J$

$\Delta K.E=-\frac{513652.48}{1000}=-513.652 KJ$

Where 1 KJ=1000 J

Change in potential energy,$\Delta P.E=mgh(h_2-h_1)$

Initially height, h1=0

Using the formula

$\Delta P.E=1100\times 9.81(40-0)$

$\Delta P.E=431640J$

$\Delta P.E=431.64KJ$