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PolarNik [594]
3 years ago
7

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.
Engineering
1 answer:
Greeley [361]3 years ago
6 0

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

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