Question

(a) Calculate the kinetic energy, in joules of a 1200 -kg automobile moving at 18 m/s .
(b) Convert this energy to calories.
(c) When the automobile brakes to a stop is the "lost" kinetic energy converted mostly to heat or to some form of potential energy?

Answer 1

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:

$K = \frac{1}{2}\cdot m \cdot v^{2}$

Where:

$m$ - Mass, measured in kilograms.

$v$ - Speed of automobile, measured in meters per second.

If $m = 1200\,kg$ and $v = 18\,\frac{m}{s}$, the kinetic energy of the automobile is:

$K = \frac{1}{2}\cdot (1200\,kg)\cdot \left(18\,\frac{m}{s} \right)^{2}$

$K = 194400\,J$

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:

$K = 194400\,J \times \left(\frac{1}{4.186}\,\frac{cal}{J} \right)$

$K = 46440.516\,cal$

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.