Question

(A) Calculate the kinetic energy, in joules, of a 1200-kg automobile moving at 18 m/s.

Answer 1

Answer:

(a) The kinetic energy in joules is 194400 J

(b) The energy in calories is 46462.72 calories

(c) When the automobile brakes to a stop, the energy will decrease to 0 J.

Explanation:

(a) Kinetic energy (K.E) is given by the formula below

$K.E = \frac{1}{2}mv^{2}$

Where K.E is the kinetic energy in Joules (J)

$m$ is the mass in kilogram (kg)

$v$ is the velocity in meter per second (m/s)

From the question,

$m =$ 1200 kg

$v =$ 18 m/s

Hence,

$K.E = \frac{1}{2}mv^{2}$ becomes

$K.E = \frac{1}{2}(1200)(18)^{2}\\K.E = (600)(324)$

$K.E = 194400$ J

This is the energy is joules

(b) To convert this energy to calories (cal)

4.184 J = 1 cal

Then, 194400 J = $x$ cal

$x = \frac{194400 \times 1}{4.184}$

$x = 46462.72$ cal

K.E = 46462.72 calories

This is the energy in calories

(c) When the automobile brakes to a stop, the energy will decrease to 0 J.