Answer:
a) The kinetic energy of the 1200-kg automobile moving at 18 meters per second is 194400 joules, b) The kinetic energy of the 1200-kg automobile moving at 18 meters per second is 46440.516 calories, c) the kinetic energy is transformed into work due to friction, which is a non-conservative force. And such work is dissipated in the form of heat.
Explanation:
a) Let be the automobile considered as particle travelling on horizontal ground, so that motion is entirely translational and whose formula for kinetic energy, measured in joules, is:
[tex]K = \frac{1}{2}\cdot m \cdot v^{2}[/tex]
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]v[/tex] - Speed of automobile, measured in meters per second.
If [tex]m = 1200\,kg[/tex] and [tex]v = 18\,\frac{m}{s}[/tex], the kinetic energy of the automobile is:
[tex]K = \frac{1}{2}\cdot (1200\,kg)\cdot \left(18\,\frac{m}{s} \right)^{2}[/tex]
[tex]K = 194400\,J[/tex]
The kinetic energy of the 1200-kg automobile moving at 18 meters per second is 194400 joules.
b) A calory equals 4.186 joules. The kinetic energy in calories is:
[tex]K = 194400\,J \times \left(\frac{1}{4.186}\,\frac{cal}{J} \right)[/tex]
[tex]K = 46440.516\,cal[/tex]
The kinetic energy of the 1200-kg automobile moving at 18 meters per second is 46440.516 calories.
c) When the automobile brakes to a stop, the kinetic energy is transformed into work due to friction, which is a non-conservative force. And such work is dissipated in the form of heat. Hence, such energy cannot be recovered. Potential energies are conservative by nature.
