The driver of a 810.0 kg car decides to double the speed from 23.6 m/s to 47.2 m/s. What effect would this have on the amount of

Question

3 years ago

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The driver of a 810.0 kg car decides to double the speed from 23.6 m/s to 47.2 m/s. What effect would this have on the amount of

work required to stop the car, that is, on the kinetic energy of the car? KEi= × 105 J KEf= × 105 J times as much work must be done to stop the car.

Mathematics

1 answer:

Setler [38] · Answer 1 · 2022-07-07T06:06:36+03:00

Answer:

KEi = 2.256×10^5 J
KEf = 9.023×10^5 J
4 times as much work

Step-by-step explanation:

The kinetic energy for a given mass and velocity is ...

KE = (1/2)mv^2 . . . . . m is mass

At its initial speed, the kinetic energy of the car is ...

KEi = (1/2)(810 kg)(23.6 m/s)^2 ≈ 2.256×10^5 J . . . . . m is meters

At its final speed, the kinetic energy of the car is ...

KEf = (1/2)(810 kg)(47.2 m/s)^2 ≈ 9.023×10^5 J

The ratio of final to initial kinetic energy is ...

(9.023×10^5)/(2.256×10^5) = 4

4 times as much work must be done to stop the car.

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You know this without computing the kinetic energy. KE is proportional to the square of speed, so when the speed doubles, the KE is multiplied by 2^2 = 4.