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

Step-by-step 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

Step-by-step 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=(1)/(2)mv_1^2`

Final kinetic energy,
`E_2=(1)/(2)mv_2^2`

`(E_1)/(E_2)=((50\ km/h)/(100\ km/h))^2`

`(E_1)/(E_2)=(1)/(4)`

`E_2=4* 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.