Question

A 1500kg car double its speed from 50km/h to 100km/h. By how many times does the kinetic energy from the car's forward motion increase

Answer 1

Answer:

Explanation:

Inital KE = (1/2) m v^2 = (1/2) * 1500 * 50^2 = 1,875,000 J  

Final KE = (1/2) * 1500 * 100^2 = 7,500,000 J  

But  ,

4 * 1875000 = 7500000

so the KE has increased by 4 times.

Answer 2

Explanation:

It is given that,

Mass of the car, m = 1500 kg

The speed of the car doubles from 50 km/h to 100 km/h

Initial kinetic energy, $E_1=\dfrac{1}{2}mv_1^2$

Final kinetic energy, $E_2=\dfrac{1}{2}mv_2^2$

$\dfrac{E_1}{E_2}=(\dfrac{50\ km/h}{100\ km/h})^2$

$\dfrac{E_1}{E_2}=\dfrac{1}{4}$

$E_2=4\times E_1$

So, the final kinetic energy from the car's forward motion becomes four times that of the initial energy. Hence, this is the required solution.