Question

A train with mass 3.3 x 107 kg starts from rest and accelerates to a speed of 42

Answer 1

Answer: The initial kinetic energy of the train is $2910.6 \times 10^{7} J$.

Explanation:

Given: Mass = $3.3 \times 10^{7} kg$

Speed = 42 m/s

Kinetic energy is the energy acquired by an object due to its motion.

Formula to calculate kinetic energy is as follows.

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

where,

m = mass of object

v = speed of object

Substitute the values into above formula as follows.

$K.E = \frac{1}{2}mv^{2}\\= \frac{1}{2} \times 3.3 \times 10^{7} kg \times (42 m/s)^{2}\\= 2910.6 \times 10^{7} kg m^{2}/s^{2} (1 J = 1 kg m^{2}/s^{2})\\= 2910.6 \times 10^{7} J$

Thus, we can conclude that the initial kinetic energy of the train is $2910.6 \times 10^{7} J$.

Answer 2

Answer:

kinetic energy of the train = 2,910.6 x 10⁷ joule

Explanation:

Given:

Mass of train = 3.3 x 10⁷ kg

Speed of train = 42 m/s

Find:

kinetic energy of the train

Computation:

kinetic energy = (1/2)(m)(v²)

kinetic energy of the train = (1/2)(3.3 x 10⁷)(42²)

kinetic energy of the train = (1/2)(3.3 x 10⁷)(1,764)

kinetic energy of the train = (3.3 x 10⁷)(882)

kinetic energy of the train = 2,910.6 x 10⁷ joule